Question -
Answer -
(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
