Combinations

We know that ..... (1)

And also n! = n(n – 1)(n – 2)…………2.1

Given that we need to find the value of

From (1)

⇒

⇒

⇒

⇒

∴ The value of is 364.

We know that ..... (1)

And also n! = n(n – 1)(n – 2)…………2.1

Given that we need to find the value of

From (1)

⇒

⇒

⇒

⇒

∴ The value of is 66.

We know that ..... (1)

And also n! = n(n – 1)(n – 2)…………2.1

Given that we need to find the value of

From (1)

⇒

⇒

We know that 0! = 1

⇒

⇒

∴ The value of is 1.

We know that ..... (1)

And also n! = n(n – 1)(n – 2)…………2.1

Given that we need to find the value of

From (1)

⇒

⇒

⇒

⇒

∴ The value of is n + 1.

We know that ..... (1)

And also n! = n(n – 1)(n – 2)…………2.1

Given that we need to find the value of

From (1)

⇒

⇒

⇒

We know that 0! = 1

⇒

⇒

⇒

∴ The value of is 31.

We know that if , then one of the following conditions need to be satisfied:

i. p = q

ii. n = p + q

From the problem we can say that,

⇒ 12≠5

So, the condition(ii) must be satisfied,

⇒ n = 12 + 5

⇒ n = 17.

∴ The value of n is 17.

We know that if , then one of the following conditions need to be satisfied:

i. p = q

ii. n = p + q

From the problem we can say that,

⇒ 4≠6

So, the condition(ii) must be satisfied,

⇒ n = 4 + 6

⇒ n = 10.

We need to find the value of

We know that ..... (1)

And also n! = n(n – 1)(n – 2)…………2.1

From (1)

⇒

⇒

⇒

⇒

∴ The value of is 66.

We know that if , then one of the following conditions need to be satisfied:

i. p = q

ii. n = p + q

From the problem we can say that,

⇒ 10≠12

So, the condition(ii) must be satisfied,

⇒ n = 10 + 12

⇒ n = 22.

We need to find the value of

We know that ..... (1)

And also n! = n(n – 1)(n – 2)…………2.1

From (1)

⇒

⇒

⇒

⇒

∴ The value of is 23.

We know that if , then one of the following conditions need to be satisfied:

i. p = q

ii. n = p + q

Let us use condition (i),

⇒ x = 2x + 3

⇒ x = – 3

We know that for a combination ⁿC_r,r≥0 and r should be an integer which is not satisfies here,

So, the condition (ii) must be satisfied,

⇒ 24 = x + 2x + 3

⇒ 3x = 21

⇒

⇒ x = 7.

∴ The value of x is 7.

We know that if , then one of the following conditions need to be satisfied:

i. p = q

ii. n = p + q

From the problem we can say that,

⇒ x≠x + 2

So, the condition(ii) must be satisfied,

⇒ 18 = x + x + 2

⇒ 2x = 16

⇒

⇒ x = 8.

∴ The value of x is 8.

We know that if , then one of the following conditions need to be satisfied:

i. p = q

ii. n = p + q

Let us use condition (i),

⇒ 3r = r + 3

⇒ 2r = 3

⇒

⇒

⇒

We know that for a combination ⁿC_r,r≥0 and r should be an integer, which is not satisfies here,

So, the condition (ii) must be satisfied,

⇒ 15 = 3r + r + 3

⇒ 4r = 12

⇒

⇒ r = 4.

∴ The value of r is 4.

Given:

⇒ ⁸C_r – ⁷C₃ = ⁷C₂

⇒ ⁸C_r = ⁷C₂ + ⁷C₃

We know that ⁿC_r + ⁿC_{r + 1} = ^{n + 1}C_{r + 1}

⇒ ⁸C_r = ^{7 + 1}C_{2 + 1}

⇒ ⁸C_r = ⁸C₃

We know that if , then one of the following conditions need to be satisfied:

i. p = q

ii. n = p + q

Let us use condition (i),

⇒ r = 3

Let us also check condition (ii),

⇒ 8 = 3 + r

⇒ r = 5

∴ The values of ‘r’ are 3 and 5.

Given:

⇒

We know that

And also n! = n(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

⇒ 5(16 – r) = 11r

⇒ 80 – 5r = 11r

⇒ 16r = 80

⇒

⇒ r = 5

∴ The value of r is 5.

Given:

⇒

We know that ,

And also n! = n(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

⇒

⇒ (n + 2)(n + 1)(n)(n – 1) = 21 × 20 × 19 × 18

Equating the corresponding terms on both sides we get,

⇒ n = 19

∴ The value of n is 19.

Given:

⇒

We know that

And also n! = n(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

⇒

⇒ (2r)(2r – 1)(2r – 2)(2r – 3) = 11 × 12 × 13 × 14

Equating the corresponding terms on both sides we get,

⇒ 2r = 14

⇒

⇒ r = 7

∴ The value of r is 7.

Given that ⁿC₄, ⁿC₅ and ⁿC₆ are in A.P.

We know that for three numbers a, b, c are in A.P, the following condition holds,

⇒ 2b = a + c

So,

⇒ 2ⁿC₅ = ⁿC₄ + ⁿC₆

Adding 2ⁿC₅ on both sides we get,

⇒ 4ⁿC₅ = ⁿC₄ + ⁿC₅ + ⁿC₅ + ⁿC₆

We know that ⁿC_r + ⁿC_{r + 1} = ^{n + 1}C_{r + 1}

⇒ 4ⁿC₅ = ^{n + 1}C₅ + ^{n + 1}C₆

⇒ 4ⁿC₅ = ^{n + 2}C₆

We know that ,

And also n! = n(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

⇒ 24(n – 4) = n² + 2n + n + 2

⇒ 24n – 96 = n² + 3n + 2

⇒ n² – 21n + 98 = 0

⇒ n² – 14n – 7n + 98 = 0

⇒ n(n – 14) – 7(n – 14) = 0

⇒ (n – 7)(n – 14) = 0

⇒ n – 7 = 0 or n – 14 = 0

⇒ n = 7 or n = 14

∴ The values of n are 7 and 14.

Given:

⇒

We know that

And also n! = n(n – 1)(n – 2)…………2.1

⇒

⇒

⇒ 2 × (2n – 1) × 2 = 44

⇒ 2n – 1 = 11

⇒ 2n = 12

⇒

⇒ n = 6

∴ The value of n is 6.

We know that if , then one of the following conditions need to be satisfied:

iii. p = q

iv. n = p + q

From the problem we can say that,

⇒ r≠r + 2

So, the condition(ii) must be satisfied,

⇒ 16 = r + r + 2

⇒ 2r = 14

⇒

⇒ r = 7.

∴ The value of r is 7.

⇒ α = ^mC₂

We know that

And also n! = n(n – 1)(n – 2)…………2.1

⇒

⇒ ..... (1)

We need to find ^αC₂

⇒

⇒

From (1)

⇒

⇒

⇒

⇒

⇒

∴ The value of ^αC₂ is .

Let us assume the negative consecutive integers are –1, – 2,......, – (2n)

Let M be the product of the negative integers,

⇒ M = ( – 1).( – 2)( – 3)......( – 2n + 1).( – 2n)

⇒ M = ( – 1)²ⁿ(1.2.3.......(2n – 1).(2n))

⇒ M = 1.2.3......……(2n – 1).(2n)

We know that, n! = n(n – 1)(n – 2)…………2.1

⇒ M = (2n)!

⇒

⇒

∴ M is divisible by (2n)!.

Given that we need to prove .

Consider L.H.S:

We know that ⁿC_r + ⁿC_{r + 1} = ^{n + 1}C_{r + 1}

⇒ ²ⁿC_n + ²ⁿC_{n – 1} = ^{2n + 1}C_n

We know that

And also n! = n(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

⇒

⇒

⇒

⇒

= R.H.S

∴ L.H.S = R.H.S, thus proved.

Given that we need to prove:

⁴ⁿC_2n : ²ⁿC_n = [1 . 3 . 5 …. (4n – 1)] : [1.3.5….. (2n – 1)]².

Consider L.H.S:

We know that

And also n! = n(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

= R.H.S

∴ L.H.S = R.H.S, thus proved.

Given that we need to find the value of .

⇒

We know that ⁿC_r + ⁿC_{r + 1} = ^{n + 1}C_{r + 1}

⇒

⇒

⇒

⇒

⇒

We know that

And also n! = n(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

⇒

∴ The value of is 42504.

Given that we need to prove

Consider L.H.S,

We know that

And also n! = n(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

= R.H.S

∴ L.H.S = R.H.S, thus proved.

Given that we need to prove n.^{n – 1}C_{r – 1} = (n – r + 1).ⁿC_{r – 1}

Consider L.H.S,

We know that

And also n! = n(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

⇒

⇒

⇒

= R.H.S

∴ L.H.S = R.H.S, thus proved.

Given that we need to prove

Consider L.H.S,

We know that

And also n! = n(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

= R.H.S

∴ L.H.S = R.H.S, thus proved.

Given that we need to prove ⁿC_r + 2ⁿC_{r – 1} + ⁿC_{r – 2} = ^{n + 2}C_r

Consider L.H.S,

We know that ⁿC_r + ⁿC_{r + 1} = ^{n + 1}C_{r + 1}

⇒ ⁿC_r + 2ⁿC_{r – 1} + ⁿC_{r – 2} = (ⁿC_r + ⁿC_{r – 1}) + (ⁿC_{r – 1} + ⁿC_{r – 2})

⇒ ⁿC_r + 2ⁿC_{r – 1} + ⁿC_{r – 2} = ^{n + 1}C_r + ^{n + 1}C_{r – 1}

⇒ ⁿC_r + 2ⁿC_{r – 1} + ⁿC_{r – 2} = ^{n + 2}C_r

= R.H.S

∴ L.H.S = R.H.S, thus proved.

Given that we need to choose 11 players for a team out of available 15 players,

It is similar to choosing ‘r’ combinations out of ‘n’ items.

i.e., ⁿC_r ways.

Let us assume the choosing the no. of ways be N,

⇒ N = choosing 11 players out of 15 players

⇒ N = ¹⁵C₁₁

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

⇒ N = 1365 ways

∴ The total no. of ways of choosing 11 players out of 15 is 1365 ways.

Given we need to find the different boat parties of 8, consisting of 5 boys and 3 girls.

The selections of girls and boys are to be made from 25 boys and 10 girls.

Let us assume the number of ways of choosing to be N.

⇒ N = (no. of ways of choosing 5 boys) × (No. of ways of choosing 3 girls)

⇒ N = (²⁵C₅) × (¹⁰C₃)

We know that ,

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

⇒ N = (53130) × (120)

⇒ N = 6375600

∴ The total no. of different boat parties are 6375600 ways.

Given that we need to find no. of ways of choosing 5 courses out of 9 courses if 2 courses are compulsory.

This is similar to choosing 3 subjects out of the remaining 7 subjects as 2 subjects are compulsory.

Let us assume the total no. of ways of choosing courses is N.

⇒ N = No. of ways of choosing 3 subjects out of 7 subjects.

⇒ N = ⁷C₃

We know that ,

And also n! = (n)(n – 1)(n – 2)......2.1

⇒

⇒

⇒

⇒ N = 35

∴ The total no. of ways of choosing 5 subjects out of 9 subjects in which 2 are compulsory is 35 ways.

Given that we need to choose 11 players for a team out of available 16 players,

Let us assume the choosing the no. of ways be N,

⇒ N = choosing 11 players out of 16 players

⇒ N = ¹⁶C₁₁

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

⇒ N = 4368 ways

(i) It is told that two players are always included.

It is similar to selecting 9 players out of the remaining 14 players as 2 players are already selected.

Let us assume the choosing the no. of ways be N₁,

⇒ N₁ = choosing 9 players out of 14 players

⇒ N₁ = ¹⁴C₉

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

⇒ N₁ = 2002 ways

(ii) It is told that two players are always excluded.

It is similar to selecting 11 players out of the remaining 14 players as 2 players are already removed.

Let us assume the choosing the no. of ways be N₂,

⇒ N₂ = choosing 11 players out of 14 players

⇒ N₂ = ¹⁴C₁₁

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

⇒ N₂ = 364 ways

∴ The required no. of ways are 4368, 2002, 364.

Given that we need to choose 2 professors and 3 students out of 10 professors and 20 students,

Let us assume the choosing the no. of ways be N,

⇒ N = (choosing 2 professors out of 10 professors) × (choosing 3 students out of 20 students)

⇒ N = (¹⁰C₂) × (²⁰C₃)

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

⇒ N = 45 × 1140

⇒ N = 51300 ways

It is told that a particular is always included.

It is similar to selecting 1 professor and 3 students out of the remaining 9 professors and 20 students as 1 professor is already selected.

Let us assume the choosing the no. of ways be N₁,

⇒ N₁ = (choosing 1 professor out of 9 professors) × (choosing 3 students out of 20 students)

⇒ N₁ = ⁹C₁ × ²⁰C₃

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

⇒ N₁ = 9 × 1140 ways

⇒ N₁ = 10260 ways

Given that we need to choose 2 professors and 3 students out of 10 professors and 20 students,

Let us assume the choosing the no. of ways be N,

⇒ N = (choosing 2 professors out of 10 professors) × (choosing 3 students out of 20 students)

⇒ N = (¹⁰C₂) × (²⁰C₃)

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

⇒ N = 45 × 1140

⇒ N = 51300 ways

It is told that one student is always included.

It is similar to selecting 2 professors and 2 students out of remaining 10 professors and 19 students as 1 student is already selected.

Let us assume the choosing the no. of ways be N₂,

⇒ N₂ = (choosing 2 professors out of 10 professors) × (choosing 2 students out of 19 students)

⇒ N₂ = ¹⁰C₂ × ¹⁹C₂

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

⇒ N₂ = 45 × 171

⇒ N₂ = 7695 ways

Given that we need to choose 2 professors and 3 students out of 10 professors and 20 students,

Let us assume the choosing the no. of ways be N,

⇒ N = (choosing 2 professors out of 10 professors) × (choosing 3 students out of 20 students)

⇒ N = (¹⁰C₂) × (²⁰C₃)

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

⇒ N = 45 × 1140

⇒ N = 51300 ways

It is told that one student is always excluded.

It is similar to selecting 2 professors and 3 students out of remaining 10 professors and 19 students as 1 student are already removed.

Let us assume the choosing the no. of ways be N₃,

⇒ N₃ = (choosing 2 professors out of 10 professors) × (choosing 3 students out of 19 students)

⇒ N₃ = ¹⁰C₂ × ¹⁹C₃

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

⇒ N₃ = 45 × 969

⇒ N₃ = 43605 ways.

∴ The required no. of ways are 51300, 10260, 7695, 43605.

Given that we need to find the no. of ways of obtaining a product by multiplying two or more from the numbers 3, 5, 7, 11.

The following are the cases the product can be done,

i. Multiplying two numbers

ii. Multiplying three numbers

iii. Multiplying four numbers

Let us assume the total numbers of ways of the product be N

⇒ N = (no. of ways of multiplying two numbers) + (no. of ways of multiplying three numbers) + (no. of multiplying four numbers)

⇒ N = ⁴C₂ + ⁴C₃ + ⁴C₄

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

⇒ N = 6 + 4 + 1

⇒ N = 11

∴ The total number of ways of product is 11 ways.

Given that 10 students need to be selected from 12 boys and 10 girls including at least 4 boys and 4 girls.

It is also told that the two girls must be includes who won prizes last year.

The cases that satisfy these conditions are:

i. Selecting 6 boys and 4 girls (in which 2 girls are already included)

ii. Selecting 5 boys and 5 girls (in which 2 girls are already included)

iii. Selecting 4 boys and 6 girls. (in which 2 girls are already included)

Let us assume the total no. of ways of selection be N,

⇒ N = (no. of ways of selecting 6 boys and 2 girls from remaining 12 boys and 8 girls) + (no. of ways of selecting 5 boys and 3 girls from remaining 12 boys and 8 girls) + (no. of ways of selecting 4 boys and 4 girls from remaining 12 boys and 8 girls)

Since, two girls are already selected,

⇒ N = (¹²C₆ × ⁸C₂) + (¹²C₅ × ⁸C₃) + (¹²C₄ × ⁸C₄)

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

⇒ N = (924 × 28) + (792 × 56) + (495 × 70)

⇒ N = 25872 + 44352 + 34650

⇒ N = 104874

∴ The total number of ways of product is 11 ways.

Given that we need to choose 4 books out of available 10 different books,

Let us assume the choosing the no. of ways be N,

⇒ N = choosing 4 books out of 10 books

⇒ N = ¹⁰C₄

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

⇒ N = 210 ways

It is told that two books are always selected.

It is similar to selecting 2 books out of the remaining 8 books as 2 books are already selected.

Let us assume the choosing the no. of ways be N₁,

⇒ N₁ = choosing 2 books out of 8 books

⇒ N₁ = ⁸C₂

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

⇒ N₁ = 28 ways

It is told that two books are never selected.

It is similar to selecting 4 books out of remaining 8books as 2 books are already removed.

Let us assume the choosing the no. of ways be N₂,

⇒ N₂ = choosing 4 books out of 8 books

⇒ N₂ = ⁸C₄

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

⇒ N₂ = 70 ways

∴ The required no. of ways are 210, 28, 70.

Given that we have 4 officers and 8 jawans, we need to choose 6 persons with the following conditions,

i. To include exactly one officer:

ii. To include at least one officer.

(i) It is told that we need to choose 6 persons with exactly one officer.

Let us assume the no. of ways of choosing to be N.

⇒ N = (no. of ways of choosing 1 officer and 5 jawans from 4 officers and 8 jawans)

⇒ N = (no. of ways of choosing 1 officer from 4 officers) × (no. of ways of choosing 5 jawans from 8 jawans)

⇒ N = (⁴C₁) × (⁸C₅)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N = 4 × 56

⇒ N = 224 ways.

(ii) It is told we need to choose 6 persons with at least 1 officers.

Let us assume the total no. of ways be N₁

⇒ N₁ = (No. of ways of choosing 6 persons with at least one officer)

⇒ N₁ = (total no. of ways of choosing 6 persons from all 12 persons) – (no. of ways of choosing 6 persons without any officer)

⇒ N₁ = ¹²C₆ – ⁸C₆

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N₁ = 924 – 28

⇒ N₁ = 896 ways

∴ The required no. of ways are 224 and 896.

Given that we need to choose a team of 11 students with at least 5 from class XI and 5 from class XII.

It is also mentioned that each class constitutes 20 students.

There are two cases of selecting a team:

i. 6 from class XI and 5 from class XII

ii. 5 from class XI and 6 from class XII

Let us assume the total no. of ways of selecting 11 students to be N.

⇒ N = no. of ways of selecting 11 students from both classes

⇒ N = (No. of ways of selecting 6 students from class XI and 5 students from class XII) + (No. of ways of selecting 5 students from class XI and 6 students from class XII)

⇒ N = (²⁰C₆ × ²⁰C₅) + (²⁰C₅ × ²⁰C₆)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N = ((38760) × (15504)) + ((15504) × (38760))

⇒ N = 600935040 + 600935040

⇒ N = 1201870080 ways

Given that 10 questions are to be answered by part A and part B by choosing at least 4 from each part.

It is also mentioned that there are 6 questions in part A and 7 in part B.

There are 3 cases to answer 10 questions:

i. 4 from part A and 6 from part B

ii. 5 from part A and 5 from part B

iii. 6 from part A and 4 from part B

Let us assume the total no. of ways of answering 10 questions be N.

⇒ N = no. of ways of answering 10 questions from both parts

⇒ N = (No. of ways of answering 4 questions from part A and 6 from part B) + (No. of ways of answering 5 questions from part A and 5 questions from part B) + (No. of ways of answering 6 questions from part A and 4 from part B)

⇒ N = (⁶C₄ × ⁷C₆) + (⁶C₅ × ⁷C₅) + (⁶C₆ × ⁷C₄)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒

⇒ N = 105 + 126 + 35

⇒ N = 266

∴ The total no. of ways of answering 10 questions is 266 ways.

Given that student have to answer 4 questions from 5 questions.

It is also told that questions 1 and 2 are compulsory.

It is similar to answering the 2 questions out of the remaining 3 questions as 1 and 2 are compulsory.

Let us assume the no. of ways of answering the questions be N.

⇒ N = No. of ways of answering 2 questions from remaining 3 questions.

⇒ N = ³C₂

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N = 3

∴ The no. of answering the questions is 3.

Given that we need to answer 7 questions from 2 groups which consist of 6 questions each.

It is also told the candidate is permitted to answer the utmost 5 questions from any group.

The cases for answering the 7 questions:

i. 5 questions from group 1 and 2 from group 2

ii. 4 questions from group 1 and 3 from group 2

iii. 3 questions from group 1 and 4 from group 2

iv. 4 questions from group 1 and 5 from group 2

Let us assume the total no. of ways of answering 7 questions be N.

⇒ N = no. of ways of answering 7 questions from both groups

⇒ N = (No. of ways of answering 5 questions from group 1 and 2 from group 2) + (No. of ways of answering 4 questions from group 1 and 3 from group 2) + (No. of ways of answering 3 questions from group 1 and 4 from group 2) + (No. of ways of answering 2 questions from group 1 and 5 from group 2)

⇒ N = (⁶C₅ × ⁶C₂) + (⁶C₄ × ⁶C₃) + (⁶C₃ × ⁶C₄) + (⁶C₂ × ⁶C₅)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N = (6 × 15) + (15 × 20) + (20 × 15) + (15 × 6)

⇒ N = 90 + 300 + 300 + 90

⇒ N = 780

∴ The total no. of ways of choosing 7 questions is 780 ways.

Given that we need to find the no. of different straight lines that can be drawn from the 10 points in which 4 are collinear.

We know that 2 points are required to draw a line and the collinear points will lie on the same line, and only one line can be drawn by joining any two points of these collinear points.

Let us assume the no. of lines formed be N,

⇒ N = (total no. of lines formed by all 10 points) – (no. of lines formed by collinear points) + 1

Here 1 is added because only 1 line can be formed by the four collinear points.

⇒ N = ¹⁰C₂ – ⁴C₂ + 1

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N = 45 – 6 + 1

⇒ N = 40

∴ The total no. of ways of different lines formed are 40.

Given that we need to find the no. of diagonals of

A hexagon

We know that the hexagon has 6 vertices and each side and diagonal can be formed by joining two vertices of a hexagon,

We know that hexagon has 6 sides,

Let us assume the no. of diagonals of the hexagon are N,

⇒ N = (no. of lines formed on joining any two vertices) – (no. of sides of the hexagon)

⇒ N = ⁶C₂ – 6

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N = 15 – 6

⇒ N = 9

∴ The total no. of diagonals formed is 9.

a polygon of 16 sides

We have given that polygon has 16 vertices and each side and diagonal can be formed by joining two vertices of a polygon,

Let us assume the no. of diagonals of the polygon are N,

⇒ N = (no. of lines formed on joining any two vertices) – (no. of sides of the polygon)

⇒ N = ¹⁶C₂ – 16

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N = 120 – 16

⇒ N = 104

∴ The total no. of diagonals formed are 104.

Given that we need to find the no. of triangles that can be drawn from the 12 points in which 5 are collinear.

We know that 3 points are required to draw a triangle and the collinear points will lie on the same line, and no triangle can be drawn by joining any three points of these collinear points.

Let us assume the no. of triangles formed be N,

⇒ N = (total no. of triangles formed by all 12 points) – (no. of triangles formed by collinear points)

⇒ N = ¹²C₃ – ⁵C₃

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N = 220 – 10

⇒ N = 210

∴ The total no. of triangles formed are 210.

Given that we need to find no. of ways of selecting 5 persons out of 6 men and 4 women in which at least one woman is necessary.

Let us assume the no. of ways of selection be N.

⇒ N = (total no. of ways of selecting 5 persons out of all 10 persons) – (No. of ways of selecting 5 persons without any women)

⇒ N = (¹⁰C₅) – (⁶C₅)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N = 252 – 6

⇒ N = 246

∴ The total no. of choosing 5 persons with at least one woman is 246.

Given that 52 families out of 82 families have at most 2 children.

It is told that 20 families need to be selected with at least 18 families having utmost 2 children.

The following are the cases:

i. 18 families having at most 2 children and 2 from other families

ii. 19 families are having at most 2 children and 1 from other families

iii. 20 families having at most 2 children

Let us assume that the no. of ways choosing to be N.

⇒ N = (Selecting 18 families having at most 2 children and 2 from other families) + (Selecting 19 families having at most 2 children and 1 from other families) + (Selecting 20 families having at most 2 children)

⇒ N = (⁵²C₁₈ × ³⁵C₂) + (⁵²C₁₉ × ³⁵C₁) + (⁵²C₂₀)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒

⇒

∴ The no. of ways of choosing 18 families are .

Given that a group consists of 4 girls and 7 boys.

We need to select a team of 5 members with the following conditions:

i. If a team has no girl

ii. If a team has at least one boy and one girl

iii. If a team has at least 3 girls.

i. Given that we need to select a team of 5 members with no girl present in it out of 4 girls and 7 boys.

Let us assume the no. of ways of selection be N

⇒ N = (selecting 5 members out of 7 boys without any girl)

⇒ N = ⁷C₅

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N = 21 ways

The no. of ways of selecting 5 members without a girl is 21 ways.

ii. Given that we need to select team of 5 members with at least 1 boy and 1 girl.

Let us assume the no. of ways of selection be N₁.

⇒ N₁ = (Total ways of selecting 5 members out of all 11 members) – (No. of ways of selecting 5 members without any girl)

⇒ N₁ = (¹¹C₅)–(⁷C₅)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N₁ = 462–21

⇒ N₁ = 441

The no. of ways of selecting 5 members with at least 1 girl and 1 boy is 441.

iii. Given that we need to find the no. of ways to select 5 members with at least 3 girls out of 7 boys and 4 girls.

The following are the possible cases:

i. Selecting 3 girls and 2 boys

ii. Selecting 4 girls and 1 boy

Let us assume the no. of ways of selection be N₂.

⇒ N₂ = (No. of ways of selecting 3 girls and 2 boys out of 7 boys and 4 girls) + (No. of ways of selecting 4 girls and 1 boy out of 7 boys and 4 girls)

⇒ N₂ = ((⁴C₃) × (⁷C₂)) + ((⁴C₄) × (⁷C₁))

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N₂ = (4 × 21) + (1 × 7)

⇒ N₂ = 84 + 7

⇒ N₂ = 91

The no. of ways of selecting 5 members with at least 3 girls is 91.

Given that we need to select 3 persons out of 2 men and 3 women,

Let us assume the no. of ways of selecting be N,

⇒ N = selecting 3 persons out of total 5 persons

⇒ N = ⁵C₃

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

⇒ N = 10 ways

The no. of ways of selecting 3 persons out of 2 men and 3 women is 10.

(ii) It is told that 1 man and 2 women should be selected out of 2 men and 3 women.

Let us assume the no. of ways of selection be N₁,

⇒ N₁ = (selecting one man out of 2 men) × (selecting 2 women out of 3 women)

⇒ N₁ = (²C₁) × (³C₂)

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

⇒ N₁ = 6 ways

The no. of ways of selecting 1 man and 2 women is 6.

i. We know that the decagon has 10 vertices and each side and diagonal can be formed by joining two vertices of a hexagon,

We know that decagon has 10 sides,

Let us assume the no. of diagonals of the hexagon are N,

⇒ N = (no. of lines formed on joining any two vertices) – (no. of sides of the hexagon)

⇒ N = ¹⁰C₂ – 10

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N = 45 – 10

⇒ N = 35

∴ The total no. of diagonals formed is 35.

ii. Given that we need to find the no. of triangles that can be drawn in a decagon.

We know that 3 points are required to draw a triangle.

We know that decagon has 10 sides

Let us assume the no. of triangles formed be N₁,

⇒ N₁ = (total no. of triangles formed by all 10 points)

⇒ N₁ = ¹⁰C₃

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N₁ = 120

∴ The total no. of ways of different lines formed are 120.

Given that we need to draw 5 cards from a deck of 52 cards.

We need to find the no. of ways that at least one of the 5 cards has to be a king.

We know that there are 4 kings present in a deck.

Let us assume the no. of ways of drawing cards be N.

⇒ N = (Total no. of ways of drawing 5 cards out of 52 cards) – (No. of ways of drawing 5 cards without a king from remaining 48 cards)

⇒ N = (⁵²C₅) – (⁴⁸C₅)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N = 2598960 – 1712304

⇒ N = 886656

∴ The no. of ways of drawing 5 cards with at least 1 king is 886656.

Given that we need to select 6 persons out of 8 persons, it is also mentioned that person B must be selected in the case of selection of A.

We need to find the no. of ways of selecting 6 persons.

The possible cases for the selection of 6 persons are:

i. selecting person A and person B and 4 others(similar to selecting 4 persons out of remaining 6 persons).

ii. selecting 6 persons other than person A and person B

Let us assume the no. of ways of selection be N.

N = (Selecting 4 persons from remaining 6 persons) + (Selecting 6 persons leaving Person A and B)

N = (⁶C₄) + (⁶C₆)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N = 15 + 1

⇒ N = 16

∴ The no. of ways of selecting 6 persons is 16.

Given that we need to select a team consisting of 3 boys and 3 girls out of 5 boys and 4 girls.

Let us assume the no. of ways of selection be N.

⇒ N = (Selecting 3 boys out of 5 boys) × (selecting 3 girls out of 4 girls)

⇒ N = (⁵C₃) × (⁴C₃)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N = (10) × (4)

⇒ N = 40

∴ The no. of ways of selecting 3 boys and 3 girls is 40.

Given that we need to select the 3 red, 3 white, and 3 blue balls out of 6 red, 5 white and 5 blue balls.

Let us assume the no. of ways of selection be N.

⇒ N = (no. of ways of selection of 3 red balls out of 6 red balls) × (no. of ways of selection of 3 white balls out of 5 white balls) × (no. of ways of selection of 3 blue balls out of 5 blue balls)

⇒ N = (⁶C₃) × (⁵C₃) × (⁵C₃)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N = 20 × 10 × 10

⇒ N = 2000

∴ The no. of ways of selection is 2000.

Given that 5 cards are drawn out of 52 cards. We know that there are 4 aces present in a deck of 52 cards.

We need to find the no. of ways of drawing 5 cards with exactly one ace out of a deck of 52 cards.

Let us assume the no. of ways of drawing cards be N.

⇒ N = (Drawing 1 ace out of 4 aces) × (Drawing 4 cards out of remaining 48 cards)

⇒ N = (⁴C₁) × (⁴⁸C₄)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N = 4 × 194580

⇒ N = 778320

∴ The no. of ways of drawing 5 cards with exactly one ace is 778320.

Given that we need to select 11 players out of available 17 players in which 5 players are bowlers.

It is also mentioned that each team should contain exactly 4 bowlers.

Let us assume the no. of ways of selection be N.

⇒ N = (no. of ways of selecting 4 bowlers out of 5) × (no. of ways of selecting 7 players from remaining 12 players)

⇒ N = (⁵C₄) × (¹²C₇)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N = 5 × 792

⇒ N = 3960

Given that we need to draw 2 black and 3 red balls from a bag of 5 black and 6 red balls.

Let us assume the no. of ways of drawing be N.

⇒ N = (no. of ways of selecting 2 black balls from 5 black balls) × (no. of ways of selecting 3 red balls from 6 red balls)

⇒ N = (⁵C₂) × (⁶C₃)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N = 10 × 20

⇒ N = 200

∴ The no. of ways of drawing 2 black and 3 red balls is 200.

Given that a student needs to choose 5 courses out of 9 courses. It is also told 2 courses are compulsory to the student.

It is similar to choosing 3 courses of remaining 7 courses since 2 courses are already chosen.

Let us assume the no. of ways of choosing to be N,

⇒ N = (no. of ways of choosing 3 courses out of remaining 7 courses)

⇒ N = ⁷C₃

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N = 35

∴ The no. of ways of choosing 5 courses is 35.

Given that we need to select 7 members out of 9 boys and 4 girls by following the conditions:

i. exactly 3 girls

ii. at least 3 girls

iii. at most 3 girls.

i. It is told we need to select 7 members out of 9 boys and 4 girls with exactly 3 girls.

Let us assume the no. of ways of selecting is N.

⇒ N = (no. of ways of selecting 3 girls out of 4 girls) × (no. of ways of selecting 4 boys out of 9 boys)

⇒ N = (⁴C₃) × (⁹C₄)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N = 4 × 126

⇒ N = 504

The no. of ways of selecting 7 members with exactly 3 girls is 504.

Given that we need to select 7 members out of 9 boys and 4 girls by following the conditions:

i. exactly 3 girls

ii. at least 3 girls

iii. at most 3 girls.

It is told we need to select 7 members out of 9 boys and 4 girls with at least 3 girls.

The possible cases are the following:

i. Selecting 3 girls and 4 boys

ii. Selecting 4 girls and 3 boys

Let us assume the no. of ways of selecting is N₁.

⇒ N₁ = ((no. of ways of selecting 3 girls out of 4 girls) × (no. of ways of selecting 4 boys out of 9 boys)) × ((no. of ways of selecting 4 girls out of 4 girls) × (no. of ways of selecting 3 boys out of 9 boys))

⇒ N₁ = ((⁴C₃) × (⁹C₄)) + ((⁴C₄) × (⁹C₃))

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N₁ = (4 × 126) + (1 × 84)

⇒ N₁ = 504 + 84

⇒ N₁ = 588

The no. of ways of selecting 7 members with at least 3 girls is 588.

Given that we need to select 7 members out of 9 boys and 4 girls by following the conditions:

i. exactly 3 girls

ii. at least 3 girls

iii. at most 3 girls.

It is told we need to select 7 members out of 9 boys and 4 girls with at most 3 girls.

The possible cases are the following:

i. Selecting 3 girls and 4 boys

ii. Selecting 2 girls and 5 boys

iii. Selecting 1 girl and 6 boys

iv. Selecting 7 boys

Let us assume the no. of ways of selecting is N₂.

⇒ N₂ = ((no. of ways of selecting 3 girls out of 4 girls) × (no. of ways of selecting 4 boys out of 9 boys)) × ((no. of ways of selecting 2 girls out of 4 girls) × (no. of ways of selecting 5 boys out of 9 boys)) × ((no. of ways of selecting 1 girls out of 4 girls) × (no. of ways of selecting 6 boys out of 9 boys))

⇒ N₂ = ((⁴C₃) × (⁹C₄)) + ((⁴C₂) × (⁹C₅)) + ((⁴C₁) × (⁹C₆)) + (⁹C₇)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N₂ = (4 × 126) + (6 × 126) + (4 × 84) + (36)

⇒ N₂ = 504 + 756 + 336 + 36

⇒ N₂ = 1632

The no. of ways of selecting 7 members with at most 3 girls is 1632.

Given that a student needs to answer 8 questions out of 12 questions in which 5 from part I and 7 from part II.

It is also told that student needs to answer at least 3 questions from each part.

The possible cases are the following:

i. 3 from part I and 5 from part II

ii. 4 from part I and 4 from part II

iii. 5 from part I and 3 from part II

Let us assume the no. of ways of answering be N.

⇒ N = (no. of ways of answering 3 questions from part I and 5 from part II) + (no. of ways of answering 4 questions from part I and 4 from part II) + (no. of ways of answering 5 questions from part I and 3 from part II)

⇒ N = ((⁵C₃) × (⁷C₅)) + ((⁵C₄) × (⁷C₄)) + ((⁵C₅) × (⁷C₃))

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N = (10 × 21) + (5 × 35) + (35)

⇒ N = 420

∴ The no. of ways of answering the question paper is 420.

We know that parallelogram has 4 lines in which 2 sides are parallel to each other which means 2 pairs of lines are parallel lines

It is told that parallelogram is cut by two sets of m lines parallel to its sides.

This means there will be two sets of (m + 2) lines parallel to each other.

We need two sets of parallel lines to form a parallelogram in which the lines need to be chosen from these two sets of (m + 2) parallel lines.

Let us assume that the no. of parallelograms formed be N.

⇒ N = (choosing 2 parallel lines from (m + 2) parallel lines which are parallel to one side) × (choosing 2 parallel lines from (m + 2) parallel lines which are parallel to the side which is not parallel to the first side)

⇒ N = (^{m + 2}C₂) × (^{m + 2}C₂)

⇒ N = (^{m + 2}C₂)²

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒

∴ The total no. of parallelograms formed are .

Given that we need to find the no. of different straight lines that can be drawn from the 18 points in which 5 are collinear.

We know that 2 points are required to draw a line and the collinear points will lie on the same line, and only one line can be drawn by joining any two points of these collinear points.

Let us assume the no. of lines formed be N,

⇒ N = (total no. of lines formed by all 18 points) – (no. of lines formed by collinear points) + 1

Here 1 is added because only 1 line can be formed by the four collinear points.

⇒ N = ¹⁸C₂ – ⁵C₂ + 1

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N = 153 – 10 + 1

⇒ N = 144

∴ The total no. of ways of different lines formed are 144.

Given that we need to find the no. of triangles that can be drawn from the 18 points in which 5 are collinear.

We know that 3 points are required to draw a triangle and the collinear points will lie on the same line, and no triangle can be drawn by joining any three points of these collinear points.

Let us assume the no. of triangles formed be N,

⇒ N = (total no. of triangles formed by all 18 points) – (no. of triangles formed by collinear points)

⇒ N = ¹⁸C₃ – ⁵C₃

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N = 816 – 10

⇒ N = 806

∴ The total no. of triangles formed are 806.

Given that we need to find the no. of words formed by 2 vowels and 3 consonants which were taken from 5 vowels and 17 consonants.

Let us find the no. of ways of choosing 2 vowels and 3 consonants and assume it to be N₁.

⇒ N₁ = (No. of ways of choosing 2 vowels from 5 vowels) × (No. of ways of choosing 3 consonants from 17 consonants)

⇒ N₁ = (⁵C₂) × (¹⁷C₃)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N₁ = 10 × 680

⇒ N₁ = 6800

Now we need to find the no. of words that can be formed by 2 vowels and 3 consonants.

Now we need to arrange the chosen 5 letters. Since every letter differs from other.

The arrangement is similar to that of arranging n people in n places which are n! ways to arrange. So, the total no. of words that can be formed is 5!.

Let us the total no. of words formed be N.

⇒ N = N₁ × 5!

⇒ N = 6800 × 120

⇒ N = 816000

∴ The no. of words that can be formed containing 2 vowels and 3 consonants are 816000.

Given that 5 persons need to be selected from 10 person P₁, P₂, P₃,……P₁₀.

It is also told that P₁ should be present and P₄ and P₅ should not be present.

It is similar to choosing 4 persons from remaining 7 persons as P₁ is selected and P₄ and P₅ are already removed.

Let us first find the no. of ways to choose persons and assume it to be N₁.

⇒ N₁ = Selecting 4 persons from remaining 7 persons

⇒ N₁ = ⁷C₄

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N₁ = 35

Now we need to arrange the chosen 5 people. Since 1 person differs from other.

The arrangement is similar to that of arranging n people in n places which are n! Ways to arrange. So, the persons can be arranged in 5! Ways.

Let us assume the total possible arrangements be N.

⇒ N = N₁ × 5!

⇒ N = 35 × 120

⇒ N = 4200

∴ The total no. of possible arrangement can be done is 4200.

Given that we need to find the no. of words formed by 4 letters which were taken from word ‘MONDAY.’

Let us find the no. of ways of choosing 4 letters and assume it to be N₁.

⇒ N₁ = (No. of ways of choosing 4 letters from MONDAY)

⇒ N₁ = (⁶C₄)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N₁ = 15

Now we need to find the no. of words that can be formed by 4 letters.

Now we need to arrange the chosen 4 letters. Since 1 person differs from other.

The arrangement is similar to that of arranging n people in n places which are n! ways to arrange. So, the total no. of words that can be formed is 4!.

Let us the total no. of words formed be N.

⇒ N = N₁ × 4!

⇒ N = 15 × 24

⇒ N = 360

∴ The no. of words that can be formed by 4 letters of MONDAY is 360.

Given that we need to find the no. of words formed by all letters of MONDAY.

Now we need to arrange the 6 letters. Since every letter differs from other.

The arrangement is similar to that of arranging n people in n places which are n! ways to arrange. So, the total no. of words that can be formed is 6!.

Let us the total no. of words formed be N.

⇒ N = 6!

⇒ N = 360

∴ The no. of words that can be formed by 6 letters of MONDAY is 360.

Given that we need to find the no. of words formed by all letters from MONDAY in which the first letter should be a vowel.

In MONDAY the vowels are O and A. We need to choose one vowel from these 2 vowels for the first place of the word.

Let us find the no. of ways of choosing vowel and assume it to be N₁.

⇒ N₁ = (No. of ways of choosing a vowel from 2 vowels)

⇒ N₁ = (²C₁)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N₁ = 2

Now we need to find the no. of words that can be formed by remaining 5 letters.

Now we need to arrange the remaining 5 letters. Since every letter differs from other. The arrangement is similar to that of arranging n people in n places which are n! ways to arrange. So, the total no. of words that can be formed is 5!.

Let us the total no. of words formed be N.

⇒ N = N₁ × 5!

⇒ N = 2 × 120

⇒ N = 240

∴ The no. of words that can be formed by all letters of MONDAY in which the first letter is a vowel is 240.

Given that we need to find the no. of permutations formed by r things which were taken from n distinct things in which 3 particular things must occur together.

Here, it is clear that 3 things are already selected and we need to choose (r – 3) things from the remaining (n – 3) things.

Let us find the no. of ways of choosing (r – 3) things and assume it to be N₁.

⇒ N₁ = (No. of ways of choosing (r – 3) things from remaining (n – 3) things)

⇒ N₁ = ^{n – 3}C_{r – 3}

Now we need to find the no. of permutations than can be formed using 3 things which are together.

Now we need to arrange the chosen 3 things. Since every thing differs from other. The arrangement is similar to that of arranging n people in n places which are n! ways to arrange. So, the total no. of words that can be formed is 3!.

Now let us assume the together things as a single thing this gives us total (r – 2) things which were present now.

Now, we need to arrange these (r – 2) things. Since every thing differs from other. The arrangement is similar to that of arranging n people in n places which are n! ways to arrange. So, the total no. of words that can be formed is (r – 2)!.

Let us the total no. of words formed be N.

⇒ N = N₁ × 3! × (r – 2)!

⇒ N = ^{n – 3}C_{r – 3} × 3! × (r – 2)!

∴ The no. of permutations that can be formed by r things which are chosen from n things in which 3 things are always together is ^{n – 3}C_{r – 3} × 3! × (r – 2)!.

Given the word is INVOLUTE. We have 4 vowels namely I,O,U,E, and consonants namely N,V,L,T.

We need to find the no. of words that can be formed using 3 vowels and 2 consonants which were chosen from the letters of involute.

Let us find the no. of ways of choosing 3 vowels and 2 consonants and assume it to be N₁.

⇒ N₁ = (No. of ways of choosing 3 vowels from 4 vowels) × (No. of ways of choosing 2 consonants from 4 consonants)

⇒ N₁ = (⁴C₃) × (⁴C₂)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N₁ = 4 × 6

⇒ N₁ = 24

Now we need to find the no. of words that can be formed by 3 vowels and 2 consonants.

Now we need to arrange the chosen 5 letters. Since every letter differs from other. The arrangement is similar to that of arranging n people in n places which are n! ways to arrange. So, the total no. of words that can be formed is 5!.

Let us the total no. of words formed be N.

⇒ N = N₁ × 5!

⇒ N = 24 × 120

⇒ N = 2880

∴ The no. of words that can be formed containing 3 vowels and 2 consonants chosen from INVOLUTE is 2880.

Given that we need to find the no. of permutations formed by r things which were taken from n distinct things in which 2 specified things must occur together.

Here, it is clear that 2 things are already selected and we need to choose (r – 2) things from the remaining (n – 2) things.

Let us find the no. of ways of choosing (r – 2) things and assume it to be N₁.

⇒ N₁ = (No. of ways of choosing (r – 2) things from remaining (n – 2) things)

⇒ N₁ = ^{n – 2}C_{r – 2}

Now we need to find the no. of permutations than can be formed using 2 things which are together.

Now we need to arrange the chosen 2 things. Since every thing differs from other. The arrangement is similar to that of arranging n people in n places which are n! ways to arrange. So, the total no. of words that can be formed is 2!.

Now let us assume the together things as a single thing this gives us total (r – 1) things which were present now.

Now, we need to arrange these (r – 1) things. Since every thing differs from other. The arrangement is similar to that of arranging n people in n places which are n! ways to arrange. So, the total no. of words that can be formed is (r – 1)!.

Let us the total no. of words formed be N.

⇒ N = N₁ × 2! × (r – 1)!

⇒ N = ^{n – 2}C_{r – 2} × 2 × (r – 1)!

∴ The no. of permutations that can be formed by r things which are chosen from n things in which 3 things are always together are ^{n – 2}C_{r – 2} × 2 × (r – 1)!.

Given the word is PROPORTION. The letters present in it are:

P: 2 in number

R: 2 in number

O: 3 in number

T: 1 in number

I: 1 in number

N: 1 in number

a. We need to find the no. of ways of selecting 4 letters from the word proportion:

The possible cases are the following:

i. 4 distinct letters

ii. 2 alike letters and 2 distinct letters.

iii. 2 alike letters of one type and 2 alike letters of another type

iv. 3 alike letters and 1 distinct letter

i. There are 6 different letters from which we need to select 4 letters. Let us assume no. of ways of selection be N₁

⇒ N₁ = no. of ways of selecting 4 letters from 6 letters

⇒ N₁ = ⁶C₄

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N₁ = 15

ii. There are 3 letters which occurred more than once. So, we need to select 1 letter from these 3 and 2 distinct letters from the remaining 5 distinct letters. Let us assume no. of ways of selection be N₂

⇒ N₂ = (no. of ways of selecting 2 alike letters from the 3 types of alike letters) × (no. of ways of selecting 2 distinct letters from remaining 5 distinct letters)

⇒ N₂ = (³C₁) × (⁵C₂)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N₂ = 3 × 10

⇒ N₂ = 30

iii. There are 3 letters which occurred more than once from which we need to select 2. Let us assume no. of ways of selection be N₃

⇒ N₃ = no. of ways of selecting 2 alike letters of one type and 2 alike letters of another type

⇒ N₃ = ³C₂

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N₃ = 3

iv. There is only 1 letter which occurred thrice, and 1 letter needs to be selected from the remaining 5 distinct letters.

Let us assume no. of ways of selection be N₄

⇒ N₄ = no. of ways of selecting 1 letter from 5 letters

⇒ N₄ = ⁵C₁

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N₄ = 5

Total no. of ways of selection = N₁ + N₂ + N₃ + N₄

Total no. of ways of selection = 15 + 30 + 3 + 5

Total no. of ways of selection = 53

b. We need to find the no. of ways of arranging 4 letters from the word proportion:

The possible cases are the following:

i. 4 distinct letters

ii. 2 alike letters and 2 distinct letters.

iii. 2 alike letters of one type and 2 alike letters of another type

iv. 3 alike letters and 1 distinct letter

i. Now we need to arrange the chosen 4 different letters. Since every word differs from other. The arrangement is similar to that of arranging n people in n places which are n! ways to arrange. So, the total no. of arrangements that can be made is 4!.

Let us assume no. of ways of arrangement be N₅.

⇒ N₅ = N₁ × 4!

⇒ N₅ = 15 × 24

⇒ N₅ = 360

ii. Now we need to arrange the chosen 2 different letters and 2 alike letters. The arrangement is similar to that of arranging n people in n places in which r are similar which are ways to arrange. So, the total no. of arrangements that can be made are .

Let us assume no. of ways of arrangement be N₆.

⇒ N₆ = N₂ ×

⇒ N₆ = 30 × 4 × 3

⇒ N₆ = 360

iii. Now we need to arrange the chosen 2 alike letters of one type and 2 alike letters of another type. The arrangement is similar to that of arranging n people in n places in which r are similar of one type and m are similar of another type which are ways too arrange. So, the total no. of arrangements that can be made are .

Let us assume no. of ways of arrangement be N₇.

⇒ N₇ = N₃ ×

⇒

⇒ N₇ = 18

iv. Now we need to arrange the chosen 1 different letters and 3 alike letters. The arrangement is similar to that of arranging n people in n places in which r are similar which are ways to arrange. So, the total no. of arrangements that can be made are .

Let us assume no. of ways of arrangement be N₈.

⇒ N₈ = N₄ ×

⇒ N₈ = 5 × 4

⇒ N₈ = 20

Total no. of ways of arrangement = N₅ + N₆ + N₇ + N₈

Total no. of ways of arrangement = 360 + 360 + 18 + 20

Total no. of ways of arrangement = 758

Given the word is MORADABAD. The letters present in it are:

M: 1 in number

O: 1 in number

R: 1 in number

A: 3 in number

D: 2 in number

B: 1 in number

a. We need to find the no. of words formed by 4 letters from the word MORADABAD:

The possible cases are the following:

i. 4 distinct letters

ii. 2 alike letters and 2 distinct letters.

iii. 2 alike letters of one type and 2 alike letters of another type

iv. 3 alike letters and 1 distinct letter

i. There are 6 different letters from which we need to select 4 letters. Let us assume no. of ways of selection be N₁

⇒ N₁ = no. of ways of selecting 4 letters from 6 letters

⇒ N₁ = ⁶C₄

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N₁ = 15

Now we need to arrange the chosen 4 different letters. Since every word differs from other. The arrangement is similar to that of arranging n people in n places which are n! ways to arrange. So, the total no. of arrangements that can be made is 4!.

Let us assume no. of ways of arrangement be N₂.

⇒ N₂ = N₁ × 4!

⇒ N₂ = 15 × 24

⇒ N₂ = 360

ii. There are 2 letters which occurred more than once. So, we need to select 1 letter from these 2 and 2 distinct letters from the remaining 5 distinct letters. Let us assume no. of ways of selection be N₃

⇒ N₃ = (no. of ways of selecting 2 alike letters from the 2 types of alike letters) × (no. of ways of selecting 2 distinct letters from remaining 5 distinct letters)

⇒ N₃ = (²C₁) × (⁵C₂)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N₃ = 2 × 10

⇒ N₃ = 20

Now we need to arrange the chosen 2 different letters and 2 alike letters. The arrangement is similar to that of arranging n people in n places in which r are similar which are ways to arrange. So, the total no. of arrangements that can be made are .

Let us assume no. of ways of arrangement be N₄.

⇒ N₄ = N₃ ×

⇒ N₄ = 20 × 4 × 3

⇒ N₄ = 240

iii. There are 2 letters which occurred more than once from which we need to select 2. Let us assume no. of ways of selection be N₅

⇒ N₅ = no. of ways of selecting 2 alike letters of one type and 2 alike letters of another type

⇒ N₅ = ²C₂

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N₅ = 1

Now we need to arrange the chosen 2 alike letters of one type and 2 alike letters of another type. The arrangement is similar to that of arranging n people in n places in which r are similar of one type and m are similar of another type which are ways too arrange. So, the total no. of arrangements that can be made are .

Let us assume no. of ways of arrangement be N₆.

⇒ N₆ = N₅ ×

⇒

⇒ N₆ = 6

iv. There is only 1 letter which occurred thrice, and 1 letter needs to be selected from the remaining 5 distinct letters.

Let us assume no. of ways of selection be N₇

⇒ N₇ = no. of ways of selecting 1 letter from 5 letters

⇒ N₇ = ⁵C₁

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N₇ = 5

Now we need to arrange the chosen 1 different letters and 3 alike letters. The arrangement is similar to that of arranging n people in n places in which r are similar which are ways to arrange. So, the total no. of arrangements that can be made are .

Let us assume no. of ways of arrangements be N₈.

⇒ N₈ = N₇ ×

⇒ N₈ = 5 × 4

⇒ N₈ = 20

Total no. of ways of words formed = N₂ + N₄ + N₆ + N₈

Total no. of ways of words formed = 360 + 240 + 6 + 20

Total no. of ways of words formed = 626

Given that we need to accommodate 21 guests to two round tables which can accommodate 15 and 6 persons each.

We need to select 6 members first and arrange 6 and 15 members accordingly in the respective tables.

Let us assume the no. of ways of choosing 6 members to be N₁

⇒ N₁ = No. of ways of choosing 6 members out of 21 guests.

⇒ N₁ = ²¹C₆

Now we need to arrange the 6 members in a round table. By fixing a guest at a single seat, We arrange the remaining 5 members. The arrangement is similar to that of arranging n people in n places which are n! ways to arrange. So, the total no. of arrangements that can be made is 5!.

Now we need to arrange the 15 members in a round table. By fixing a guest at a single seat, We arrange the remaining 5 members. The arrangement is similar to that of arranging n people in n places which are n! ways to arrange. So, the total no. of arrangements that can be made is 14!.

Let us assume total no. of ways of arranging the guests in the table be N

⇒ N = N₁ × 5! × 14!

⇒ N = ²¹C₆ × 5! × 14!

∴ The no. of ways of accommodating guests is ²¹C₆ × 5! × 14!.

Given the word is EXAMINATION. The letters present in it are:

E: 1 in number

X: 1 in number

A: 2 in number

M: 1 in number

I: 2 in number

N: 2 in number

T: 1 in number

O: I in number

a. We need to find the no. of words formed by 4 letters from the word EXAMINATION:

The possible cases are the following:

i. 4 distinct letters

ii. 2 alike letters and 2 distinct letters.

iii. 2 alike letters of one type and 2 alike letters of another type

i. There are 8 different letters from which we need to select 4 letters. Let us assume no. of ways of selection be N₁

⇒ N₁ = no. of ways of selecting 4 letters from 8 letters

⇒ N₁ = ⁸C₄

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N₁ = 70

Now we need to arrange the chosen 4 different letters. Since every word differs from other. The arrangement is similar to that of arranging n people in n places which are n! ways to arrange. So, the total no. of arrangements that can be made is 4!.

Let us assume no. of ways of arrangement be N₂.

⇒ N₂ = N₁ × 4!

⇒ N₂ = 70 × 24

⇒ N₂ = 1680

ii. There are 3 letters which occurred more than once. So, we need to select 1 letter from these 2 and 2 distinct letters from the remaining 7 distinct letters. Let us assume no. of ways of selection be N₃

⇒ N₃ = (no. of ways of selecting 2 alike letters from the 3 types of alike letters) × (no. of ways of selecting 2 distinct letters from remaining 5 distinct letters)

⇒ N₃ = (³C₁) × (⁷C₂)

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N₃ = 3 × 21

⇒ N₃ = 63

Now we need to arrange the chosen 2 different letters and 2 alike letters. The arrangement is similar to that of arranging n people in n places in which r are similar which are ways to arrange. So, the total no. of arrangements that can be made are .

Let us assume no. of ways of arrangement be N₄.

⇒ N₄ = N₃ ×

⇒ N₄ = 63 × 4 × 3

⇒ N₄ = 756

iii. There are 3 letters which occurred more than once from which we need to select 2. Let us assume no. of ways of selection be N₅

⇒ N₅ = no. of ways of selecting 2 alike letters of one type and 2 alike letters of other type

⇒ N₅ = ³C₂

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N₅ = 3

Now we need to arrange the chosen 2 alike letters of one type and 2 alike letters of another type. The arrangement is similar to that of arranging n people in n places in which r are similar of one type and m are similar of another type which are ways too arrange. So, the total no. of arrangements that can be made are .

Let us assume no. of ways of arrangement be N₆.

⇒ N₆ = N₅ ×

⇒

⇒ N₆ = 18

Total no. of combinations = N₁ + N₃ + N₅

Total no. of combinations = 70 + 63 + 3

Total no. of combinations = 136

Total no. of ways of permutations = N₂ + N₄ + N₆

Total no. of ways of permutations = 1680 + 756 + 18

Total no. of ways of permutations = 2454

Given that 16 persons need to be seated along two sides of a long table with 8 persons on each side.

It is also told that 4 persons sit on a particular side and 2 on the other side.

We need to choose 4 members to sit on one side from the remaining 10 members as the 6 members are already fixed about their seatings and arrange 8 members on both sides accordingly.

Let us first find the no. of ways to choose 4 members and assume it to be N₁.

⇒ N₁ = Selecting 4 members from remaining 7 person members

⇒ N₁ = ¹⁰C₄

We know that ,

And also n! = (n)(n – 1)......2.1

⇒

⇒

⇒

⇒ N_{1 =} 210

Now we need to arrange the chosen 8 members. Since 1 person differs from other.

The arrangement is similar to that of arranging n people in n places which are n! Ways to arrange. So, the persons can be arranged in 8! Ways.

This will be the same for both the tables.

Let us assume the total possible arrangements be N.

⇒ N = N₁ × 8! × 8!

⇒ N = 210 × 8! × 8!

∴ The total no. of ways of seating arrangements can be done 210 × 8! × 8!.

We know:

ⁿC_r + ⁿC_r-1 = ⁿ⁺¹C_r ⇒ (1)

^n+rC_r = ⁿC₀ + ⁿ⁺¹C₁ + ⁿ⁺²C₂ + ⁿ⁺³C₃ + . . . . . . + ^n+mC_n+1

^n+rC_r = ⁿ⁺¹C₀ + ⁿ⁺¹C₁ + ⁿ⁺²C₂ + ⁿ⁺³C₃ + . . . . . . + ^n+mC_n+1 ⇒ (ⁿC₀ = ⁿ⁺¹C₀)

Using equation (1),

^n+rC_r = ⁿ⁺²C₁ + ⁿ⁺²C₂ + ⁿ⁺³C₃ + . . . . . . + ^n+mC_n+1

^n+rC_r = ⁿ⁺³C₂ + ⁿ⁺³C₃ + . . . . . . + ^n+mC_n+1

Proceeding in the same way :

^n+rC_r = ^n+mC_m-1 + ^n+mC_m = ^n+m+1C_n+1

^n+rC_r = ^n+m+1C_n+1

³⁵C_n+7 = ³⁵C_4n-2

n + 7 + 4n – 2 = 35 ( ⁿC_{x =}ⁿC_y ⇒ n = x + y or x = y )

5n + 5 = 35

5n = 30

n = 6

And,

n + 7 = 4n – 2

3n = 9

n = 3

An n-sided polygon has n vertices.

By joining any two vertices of the polygon, we obtain either a side or a diagonal of the polygon.

Number of line segments obtained by joining the vertices of an n-sided polygon if we take two vertices at a time, number of

Ways of selection 2 out of n = ⁿC₂

Out of these lines, n lines are sides of the polygon.

Number of diagonals of the polygon =

ⁿC₂ – n – n

ⁿC_r+1 + ⁿC_r-1 + 2 ⁿC_r

= (ⁿC_r+1 + ⁿC_r) + (ⁿC_r + ⁿC_r-1)

⇒ (ⁿC_r + ⁿC_r-1 = ⁿ⁺¹C_r)

= ⁿ⁺¹C_r+1 + ⁿ⁺¹C_r

⇒ (ⁿC_r + ⁿC_r-1 = ⁿ⁺¹C_r)

= ⁿ⁺²C_r+1

We know,

ⁿC_r + ⁿC_r-1 = ⁿ⁺¹C_r

Now, we have,

^56-rC₃ + ⁵⁰C₄

= ⁵⁵C₃ + ⁵⁴C₃ + ⁵³C₃ + ⁵²C₃ + ⁵¹C₃ + ⁵⁰C₃ + ⁵⁰C₄

= ⁵⁵C₃ + ⁵⁴C₃ + ⁵³C₃ + ⁵²C₃ + ⁵¹C₃ + ⁵¹C₄

= ⁵⁵C₃ + ⁵⁴C₃ + ⁵³C₃ + ⁵²C₃ + ⁵²C₄

= ⁵⁵C₃ + ⁵⁴C₃ + ⁵³C₃ + ⁵³C₄

= ⁵⁵C₃ + ⁵⁴C₃ + ⁵⁴C₄

= ⁵⁵C₃ + ⁵⁵C₄

= ⁵⁶C₄

Total number of ways in which the letters can be put = 3! = 6

Suppose, out of the three letters, one has been put in the correct envelope.

This can be done in ³C₁ ways. (3 ways)

Now, out of three, if two letters have been put in the current envelope, then the last one has been put in

the correct envelope as well. This can be done in ³C₃ ways. (1 way)

Number of ways = 3 + 1 = 4

Number of ways in which no letter is put in correct envelope = 6 – 4 = 2

We know that two lines are required for one point of intersection.

Number of points of intersection =

⁸C₂

= 28

A parallelogram can be formed by choosing two parallel lines from the set of four parallel lines and two

parallel lines from the set of three parallel lines.

Two parallel lines from the set of four parallel lines can be chosen in ⁴C₂ ways.

Two parallel lines from the set of three parallel lines can be chosen in ³C₂ ways.

Number of parallelograms that can be formed =

⁴C₂³C₂

= 6 3

= 18

4 white and 5 red balls are to be selected from 8 white and 10 red balls.

Required number of ways = ⁸C₄¹⁰C₅

= 70 252

= 17640

Number of groups in which 12 boys are to be divided = 3

Now 4 boys can be chosen out of 12 boys in (¹²C₄⁸C₄⁴C₄) ways.

=

These groups can be arranged in 3! ways.

Total number of ways =

= 5775

2 out of 4 vowels and 3 out of 5 consonants can be chosen in ⁴C₂⁵C₃ ways.

The total number of letters is 5. These letters can be arranged in 5! ways.

Total number of words =

⁴C₂⁵C₃ 5!

= 6 10 120

= 7200

r + r – 10 = 20 ( ⁿC_{x =}ⁿC_y ⇒ n = x + y or x = y )

2r – 10 = 20

2r = 30

r = 15

Now,

¹⁸C_r = ¹⁸C₁₅

¹⁸C₁₅ = ¹⁸C₃

¹⁸C₃

r + r + 4 = 20 ( ⁿC_{x =}ⁿC_y ⇒ n = x + y or x = y )

2r + 4 = 20

2r = 16

r = 8

Now,

^rC₃ = ⁸C₃

⁸C₃

3r + r + 3 = 15 ( ⁿC_{x =}ⁿC_y ⇒ n = x + y or x = y )

4r + 3 = 15

4r = 12

r = 3

r + 1 + r – 1 = 20 ( ⁿC_{x =}ⁿC_y ⇒ n = x + y or x = y )

2r = 20

r = 10

ⁿC_{12 =}ⁿC₈

n = 12 + 8 ( ⁿC_{x =}ⁿC_y ⇒ n = x + y or x = y )

n = 20

Now,

²²C_n = ²²C₂₀

= 231

^mC₁ = ⁿC₂

2m = n(n – 1)

n = 12 + 8 ( ⁿC_{x =}ⁿC_y ⇒ n = x + y or x = y )

n = 20

ⁿC_r + ⁿC_r+1 = ⁿ⁺¹C_x (Given)

Now, we have ⇒

ⁿC_r + ⁿC_r-1 = ⁿ⁺¹C_r ⇒ (1)

From (1) and (Given) we have,

ⁿ⁺¹C_r+1 = ⁿ⁺¹C_x

r + 1 = x ( ⁿC_{x =}ⁿC_y ⇒ n = x + y or x = y )

a² – a = 2 + 4 ( ⁿC_{x =}ⁿC_y ⇒ n = x + y or x = y )

a² – a – 6 = 0

a² – 3a – 2a – 6 = 0

a(a – 3) + 2(a – 3) = 0

(a + 2) (a – 3) = 0

a = -2 or a = 3

But, a = -2 is not possible since it’s negative.

So, a = 3.

⁵C₁ + ⁵C₂ + ⁵C₃ + ⁵C₄ + ⁵C₅

= ⁵C₁ + ⁵C₂ + ⁵C₂ + ⁵C₁ + ⁵C₅ (⁵C₁ = ⁵C₄ and ⁵C₂ = ⁵C₃)

= 2 ⁵C₁ + 2 ⁵C₂ + ⁵C₅

= 2 5 + 2 + 1

= 10 + 20 + 1

= 31

2 out of 4 vowels can be chosen in ⁴C₂ ways.

3 out of 5 consonants can be chosen in ⁵C₃ ways.

Thus, there are (⁴C₂⁵C₃) groups, each containing 2 vowels and 3 consonants.

Each group contains 5 letters that can be arranged in 5! Ways.

Required number of words =

(⁴C₂⁵C₃) 5!

= 60 120

= 7200

Number of straight lines joining 12 points if we take 2 points at a time = ¹²C₂

= 66

Number of straight lines joining 3 points if we take 2 points at a time = ³C₂

= 3

But, 3 collinear points, when joined in pairs gives only one line.

Required number of straight lines =

66 – 3 + 1

= 64

Three persons can take 5 seats in ⁵C₃ ways.

Also, 3 persons can sit in 3! ways.

Required number of ways =

⁵C₃ 3!

= 10 6

= 60

Required number of ways =

⁴C₁⁶C₄ + ⁴C₂⁶C₃ + ⁴C₃⁶C₂ + ⁴C₄⁶C₁

= 60 + 120 + 60 + 6

= 246

Number of straight lines formed by joining the 10 points if we take 2 points at a time = ¹⁰C₂

Number of straight lines formed by joining the 4 points if we take 2 points at a time = ⁴C₂

But, 4 collinear points, when joined in pairs give only one line.

Required number of straight lines =

45 – 6 + 1

= 40

4 out of 13 players are bowlers.

In other words, 9 players are not bowlers.

A team of 11 is to be selected so as to include at least 2 bowlers.

Number of ways = ⁴C₂⁹C₉ + ⁴C₃⁹C₈ + ⁴C₄⁹C₇

= 6 + 36 + 36

= 78

If set S has n elements, then C(n,k) is the number of ways of choosing k elements from S.

Thus, the number of subsets of S of all possible values is given by,

C(n,0) + C(n,1) + C(n,2) + . . . . . . . . . . + C(n,n) = 2ⁿ

Comparing the given equation with the above equation we get,

2ⁿ = 256

2ⁿ = 2⁸

n = 8

²ⁿC₂ = ¹⁶C₂

= 120

A host lady can invite 8 out of 12 people in ¹²C₈ ways.

Two out of these 12 people do not want to attend the party together.

Number of ways =

¹²C₈ - ¹⁰C₆

We need at least three points to draw a circle that passes through them.

Now, number of circles formed out of 11 points by taking three points at a time = ¹¹C₃

= 165

Number of circles formed out of 5 points by taking three points at a time = ⁵C₃

= 10

It is given that 5 points lie on once circle.

Required number of circles =

165 – 10 + 1

= 156

Number of committees that can be formed ⇒

= ⁶C₃⁴C₂

= 120

r – 6 + 3r + 1 = 43 ( ⁿC_{x =}ⁿC_y ⇒ n = x + y or x = y )

4r – 5 = 43

4r = 48

r = 12

An octagon has 8 vertices.

The number of diagonals of a polygon is given by . (Where n = number of vertices)

Number of diagonals of an octagon =

( ⁷C₀ + ⁷C₁ ) + ( ⁷C₁ + ⁷C₂ ) + ( ⁷C₂ + ⁷C₃ ) + ( ⁷C₃ + ⁷C₄ ) + ( ⁷C₄ + ⁷C₅ ) + ( ⁷C₅ + ⁷C₆ ) + ( ⁷C₆ + ⁷C₇ )

= 1 + 2⁷C₁ + 2⁷C₂ + 2⁷C₃ + 2⁷C₄ + 2⁷C₅ + 2⁷C₆ + 1

= 1 + 2⁷C₁ + 2⁷C₂ + 2⁷C₃ + 2⁷C₃ + 2⁷C₂ + 2⁷C₆ + 1 ⇒ (⁷C₃ = ⁷C₄ and ⁷C₂ = ⁷C₅)

= 2 + 2² (⁷C₁ + ⁷C₂ + ⁷C₃)

= 2 + 2² (7 + 6 + 5)

= 2 + 252

= 254

= 2⁸ – 2

5 out of 14 players are bowlers.

In other words, 9 players are not bowlers.

A team of 11 is to be selected so as to include at least 4 bowlers.

Number of ways = ⁵C₄⁹C₇ + ⁵C₅⁹C₆

= 180 + 84

= 264

Suppose there are two friends, A and B, who do not attend the party together.

If both of them do not attend the party, then the number of ways selecting 6 guests = ⁸C₆ = 28

If one of them attends the party, then the number of ways of selecting 6 guests = 2 ⁸C₅ = 112

Total number of ways = 112 + 28 = 140

ⁿ⁺¹C₃ = 2 ⁿC₂

n + 1 = 6

n = 5

A parallelogram can be formed by choosing two parallel lines from the set of four parallel lines and two

parallel lines from the set of three parallel lines.

Two parallel lines from the set of four parallel lines can be chosen in ⁴C₂ ways.

Two parallel lines from the set of three parallel lines can be chosen in ³C₂ ways.

Number of parallelograms that can be formed =

⁴C₂³C₂

= 6 3

= 18