In this section, we shall discuss some important properties of combinations C(n,r). Students who find difficulty in understanding the concepts and solving problems can make use of the PDF of solutions. Our expert faculty at BYJU’S, having vast experience in their respective fields, have designed the solutions. The problems are solved in a structured manner so that it helps students understand each step, as each step carries marks in the board exam. Students can improve their problem-solving and logical thinking abilities by using RD Sharma Class 11 Maths Solutions PDF from the links given below.
RD Sharma Solutions for Class 11 Maths Exercise 17.1 Chapter 17 – Combinations
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Access answers to RD Sharma Solutions for Class 11 Maths Exercise 17.1 Chapter 17 – Combinations
1. Evaluate the following:
(i) 14C3
(ii) 12C10
(iii) 35C35
(iv) n+1Cn
(v) 
Solution:
(i) 14C3
Let us use the formula,
nCr = n!/r!(n – r)!
So now, value of n = 14 and r = 3
nCr = 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,
nCr = n!/r!(n – r)!
So now, value of n = 12 and r = 10
nCr = 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,
nCr = n!/r!(n – r)!
So now, value of n = 35 and r = 35
nCr = n!/r!(n – r)!
35C35 = 35! / 35!(35 – 35)!
= 35! / (35! 0!) [Since, 0! = 1]
= 1
(iv) n+1Cn
Let us use the formula,
nCr = n!/r!(n – r)!
So now, value of n = n+1 and r = n
nCr = n!/r!(n – r)!
n+1Cn = (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 nCp = nCq, 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 12Cn.
Solution:
We know that if nCp = nCq, 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 12Cn,
We know the value of n so, 12Cn = 12C10
Let us use the formula,
nCr = n!/r!(n – r)!
So now, value of n = 12 and r = 10
nCr = 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 23Cn.
Solution:
We know that if nCp = nCq, 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 23Cn,
We know the value of n so, 23Cn = 23C22
Let us use the formula,
nCr = n!/r!(n – r)!
So now, value of n = 23 and r = 22
nCr = n!/r!(n – r)!
23C22 = 23! / 22!(23 – 22)!
= 23! / (22! 1!)
= [23×22!] / (22!)
= 23
∴ The value of 23C22 = 23.
5. If 24Cx = 24C2x + 3, find x.
Solution:
We know that if nCp = nCq, then one of the following conditions need to be satisfied:
(i) p = q
(ii) n = p + q
So from the question 24Cx = 24C2x + 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 nCr, 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 18Cx = 18Cx + 2, find x.
Solution:
We know that if nCp = nCq, then one of the following conditions need to be satisfied:
(i) p = q
(ii) n = p + q
So from the question 18Cx = 18Cx + 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 15C3r = 15Cr + 3, find r.
Solution:
We know that if nCp = nCq, then one of the following conditions need to be satisfied:
(i) p = q
(ii) n = p + q
So from the question 15C3r = 15Cr + 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 nCr, 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 8Cr – 7C3 = 7C2, find r.
Solution:
To find r, let us consider the given expression,
8Cr – 7C3 = 7C2
8Cr = 7C2 + 7C3
We know that nCr + nCr + 1 = n + 1Cr + 1
8Cr = 7 + 1C2 + 1
8Cr = 8C3
Now, we know that if nCp = nCq, then one of the following conditions need to be satisfied:
(i) p = q
(ii) n = p + q
So from the question 8Cr = 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 15Cr: 15Cr – 1 = 11: 5, find r.
Solution:
Given:
15Cr: 15Cr – 1 = 11: 5
15Cr / 15Cr – 1 = 11 / 5
Let us use the formula,
nCr = 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 + 2C8: n – 2P4 = 57: 16, find n.
Solution:
Given:
n + 2C8: n – 2P4 = 57: 16
n + 2C8 / n – 2P4 = 57 / 16
Let us use the formula,
nCr = n!/r!(n – r)!
(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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