RD Sharma Solutions Class 11 Maths Chapter 17 Exercise 17.1 (Updated for 2024)

RD Sharma Solutions Class 11 Maths Chapter 17 Exercise 17.1: Previously we have studied the arrangement of a fixed number of objects acquiring some or all at the same time. Part or all of the elements, regardless of their arrangement, each of the different options is called a combination. Students who wish to learn by themselves and solve problems encountered in practice can use RD Sharma Solutions Class 11 Maths Chapter 17 Exercise 17.1 as reference material, which is the best resource available to any student. All solutions are created by experts in the field, taking into account the latest CBSE marking model. The exercise-based solutions for these topics are provided in PDF format, which can be easily downloaded by students.

• RD Sharma Solutions Class 11 Maths 

Download RD Sharma Solutions Class 11 Maths Chapter 17 Exercise 17.1 PDF:

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Access RD Sharma Solutions Class 11 Maths Chapter 17 Exercise 17.1

1. Evaluate the following:

(i) ¹⁴C₃

(ii) ¹²C₁₀

(iii) ³⁵C₃₅

(iv) ⁿ⁺¹C_n

(v) 

Solution:

(i) ¹⁴C₃

Let us use the formula,

ⁿC_r = n!/r!(n – r)!

So now, value of n = 14 and r = 3

ⁿC_r = n!/r!(n – r)!

¹⁴C₃ = 14! / 3!(14 – 3)!

= 14! / (3! 11!)

= [14×13×12×11!] / (3! 11!)

= [14×13×12] / (3×2)

= 14×13×2

= 364

(ii) ¹²C₁₀

Let us use the formula,

ⁿC_r = n!/r!(n – r)!

So now, value of n = 12 and r = 10

ⁿC_r = n!/r!(n – r)!

¹²C₁₀ = 12! / 10!(12 – 10)!

= 12! / (10! 2!)

= [12×11×10!] / (10! 2!)

= [12×11] / (2)

= 6×11

= 66

(iii) ³⁵C₃₅

Let us use the formula,

ⁿC_r = n!/r!(n – r)!

So now, value of n = 35 and r = 35

ⁿC_r = n!/r!(n – r)!

³⁵C₃₅ = 35! / 35!(35 – 35)!

= 35! / (35! 0!) [Since, 0! = 1]

= 1

(iv) ⁿ⁺¹C_n

Let us use the formula,

ⁿC_r = n!/r!(n – r)!

So now, value of n = n+1 and r = n

ⁿC_r = n!/r!(n – r)!

ⁿ⁺¹C_n = (n+1)! / n!(n+1 – n)!

= (n+1)! / n!(1!)

= (n + 1) / 1

= n + 1

2. If ⁿC₁₂ = ⁿC₅, find the value of n.

Solution:

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

(i) p = q

(ii) n = p + q

So from the question ⁿC₁₂ = ⁿC₅, we can say that

12 ≠ 5

So, condition (ii) must be satisfied,

n = 12 + 5

n = 17

∴ The value of n is 17.

3. If ⁿC₄ = ⁿC₆, find ¹²C_n.

Solution:

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

(i) p = q

(ii) n = p + q

So from the question ⁿC₄ = ⁿC₆, we can say that

4 ≠ 6

So, condition (ii) must be satisfied,

n = 4 + 6

n = 10

Now, we need to find ¹²C_n,

We know the value of n so, ¹²C_n = ¹²C₁₀

Let us use the formula,

ⁿC_r = n!/r!(n – r)!

So now, value of n = 12 and r = 10

ⁿC_r = n!/r!(n – r)!

¹²C₁₀ = 12! / 10!(12 – 10)!

= 12! / (10! 2!)

= [12×11×10!] / (10! 2!)

= [12×11] / (2)

= 6×11

= 66

∴ The value of ¹²C₁₀ = 66.

4. If ⁿC₁₀ = ⁿC₁₂, find ²³C_n.

Solution:

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

(i) p = q

(ii) n = p + q

So from the question ⁿC₁₀ = ⁿC₁₂, we can say that

10 ≠ 12

So, condition (ii) must be satisfied,

n = 10 + 12

n = 22

Now, we need to find ²³C_n,

We know the value of n so, ²³C_n = ²³C₂₂

Let us use the formula,

ⁿC_r = n!/r!(n – r)!

So now, value of n = 23 and r = 22

ⁿC_r = n!/r!(n – r)!

²³C₂₂ = 23! / 22!(23 – 22)!

= 23! / (22! 1!)

= [23×22!] / (22!)

= 23

∴ The value of ²³C₂₂ = 23.

5. If ²⁴C_x = ²⁴C_{2x + 3}, find x.

Solution:

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

(i) p = q

(ii) n = p + q

So from the question ²⁴C_x = ²⁴C_{2x + 3}, we can say that

Let us check for condition (i)

x = 2x + 3

2x – x = -3

x = -3

We know that for a combination ⁿC_r, r≥0, r should be a positive integer which is not satisfied here,

So, condition (ii) must be satisfied,

24 = x + 2x + 3

3x = 21

x = 21/3

x = 7

∴ The value of x is 7.

6. If ¹⁸C_x = ¹⁸C_{x + 2}, find x.

Solution:

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

(i) p = q

(ii) n = p + q

So from the question ¹⁸C_x = ¹⁸C_{x + 2}, we can say that

x ≠ x + 2

So, condition (ii) must be satisfied,

18 = x + x + 2

18 = 2x + 2

2x = 18 – 2

2x = 16

x = 16/2

= 8

∴ The value of x is 8.

7. If ¹⁵C_3r = ¹⁵C_{r + 3}, find r.

Solution:

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

(i) p = q

(ii) n = p + q

So from the question ¹⁵C_3r = ¹⁵C_{r + 3}, we can say that

Let us check for condition (i)

3r = r + 3

3r – r = 3

2r = 3

r = 3/2

We know that for a combination ⁿC_r, r≥0, r should be a positive integer which is not satisfied here,

So, condition (ii) must be satisfied,

15 = 3r + r + 3

15 – 3 = 4r

4r = 12

r = 12/4

= 3

∴ The value of r is 3.

8. If ⁸C_r – ⁷C₃ = ⁷C₂, find r.

Solution:

To find r, let us consider the given expression,

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

Now, we know that if ⁿC_p = ⁿC_q, then one of the following conditions need to be satisfied:

(i) p = q

(ii) n = p + q

So from the question ⁸C_r = ⁸C₃, we can say that

Let us check for condition (i)

r = 3

Let us also check for condition (ii)

8 = 3 + r

r = 5

∴ The values of ‘r’ are 3 and 5.

9. If ¹⁵C_r: ¹⁵C_{r – 1} = 11: 5, find r.

Solution:

Given:

¹⁵C_r: ¹⁵C_{r – 1} = 11: 5

¹⁵C_r / ¹⁵C_{r – 1} = 11 / 5

Let us use the formula,

ⁿC_r = n!/r!(n – r)!

5(16 – r) = 11r

80 – 5r = 11r

80 = 11r + 5r

16r = 80

r = 80/16

= 5

∴ The value of r is 5.

10. If ^{n + 2}C₈: ^{n – 2}P₄ = 57: 16, find n.

Solution:

Given:

^{n + 2}C₈: ^{n – 2}P₄ = 57: 16

^{n + 2}C₈ / ^{n – 2}P₄ = 57 / 16

Let us use the formula,

ⁿC_r = n!/r!(n – r)!

[(n+2)! (n-6)!] / [(n-6)! (n-2)! 8!] = 57/16

(n+2) (n+1) (n) (n-1) / 8! = 57/16

(n+2) (n+1) (n) (n-1) = (57×8!) / 16

(n+2) (n+1) (n) (n-1) = [19×3 × 8×7×6×5×4×3×2×1]/16

(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.

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FAQ: RD Sharma Solutions Class 11 Maths Chapter 17 Exercise 17.1