RD Chapter 17 Combinations Ex 17.1 Solutions
Question - 1 : - Evaluate the following:
(i) 14C3
(ii) 12C10
(iii) 35C35
(iv) n+1Cn
(v) 
Answer - 1 : -
(i) 14C3
Let us use theformula,
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 theformula,
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 theformula,
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 theformula,
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
Question - 2 : - If nC12 = nC5,find the value of n.
Answer - 2 : -
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 thequestion nC12 = nC5,we can say that
12 ≠ 5
So, the condition (ii)must be satisfied,
n = 12 + 5
n = 17
∴ The value of nis 17.
Question - 3 : - If nC4 = nC6,find 12Cn.
Answer - 3 : -
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 thequestion nC4 = nC6,we can say that
4 ≠ 6
So, the condition (ii)must be satisfied,
n = 4 + 6
n = 10
Now, we need tofind 12Cn,
We know the value of nso, 12Cn = 12C10
Let us use theformula,
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 valueof 12C10 = 66.
Question - 4 : - If nC10 = nC12,find 23Cn.
Answer - 4 : -
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 thequestion nC10 = nC12,we can say that
10 ≠ 12
So, the condition (ii)must be satisfied,
n = 10 + 12
n = 22
Now, we need tofind 23Cn,
We know the value of nso, 23Cn = 23C22
Let us use theformula,
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 valueof 23C22 = 23.
Question - 5 : - If 24Cx = 24C2x + 3,find x.
Answer - 5 : -
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 thequestion 24Cx = 24C2x + 3,we can say that
Let us check forcondition (i)
x = 2x + 3
2x – x = -3
x = -3
We know that for acombination nCr, r≥0, r should be a positive integerwhich is not satisfied here,
So, the condition (ii)must be satisfied,
24 = x + 2x + 3
3x = 21
x = 21/3
x = 7
∴ The value of xis 7.
Question - 6 : - If 18Cx = 18Cx + 2,find x.
Answer - 6 : -
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 thequestion 18Cx = 18Cx + 2,we can say that
x ≠ x + 2
So, the condition (ii)must be satisfied,
18 = x + x + 2
18 = 2x + 2
2x = 18 – 2
2x = 16
x = 16/2
= 8
∴ The value of xis 8.
Question - 7 : - If 15C3r = 15Cr + 3,find r.
Answer - 7 : -
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 thequestion 15C3r = 15Cr + 3,we can say that
Let us check forcondition (i)
3r = r + 3
3r – r = 3
2r = 3
r = 3/2
We know that for acombination nCr, r≥0, r should be a positive integerwhich is not satisfied here,
So, the condition (ii)must be satisfied,
15 = 3r + r + 3
15 – 3 = 4r
4r = 12
r = 12/4
= 3
∴ The value of ris 3.
Question - 8 : - If 8Cr – 7C3 = 7C2,find r.
Answer - 8 : -
To find r, let usconsider 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 thatif nCp = nCq, thenone of the following conditions need to be satisfied:
(i) p = q
(ii) n = p + q
So from thequestion 8Cr = 8C3,we can say that
Let us check forcondition (i)
r = 3
Let us also check forcondition (ii)
8 = 3 + r
r = 5
∴ The values of‘r’ are 3 and 5.
Question - 9 : - If 15Cr: 15Cr – 1 =11: 5, find r.
Answer - 9 : -
Given:
15Cr: 15Cr – 1 =11: 5
15Cr / 15Cr– 1 = 11 / 5
Let us use theformula,
nCr = n!/r!(n – r)!
5(16 – r) = 11r
80 – 5r = 11r
80 = 11r + 5r
16r = 80
r = 80/16
= 5
∴ The value of ris 5.
Question - 10 : - If n + 2C8: n – 2P4 =57: 16, find n.
Answer - 10 : -
Given:
n + 2C8: n – 2P4 =57: 16
n + 2C8 / n – 2P4 =57 / 16
Let us use theformula,
nCr = 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 thecorresponding terms on both sides we get,
n = 19
∴ The value of nis 19.