Permutations and Combinations 2 Question 1

1. The number of ways of choosing 10 objects out of 31 objects of which 10 are identical and the remaining 21 are distinct, is

(2019 Main, 12 April I)

(a) 2201

(b) 221

(c) 220

(d) 220+1

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Answer:

Correct Answer: 1. (c)

Solution:

  1. Given that, out of 31 objects 10 are identical and remaining 21 are distinct, so in following ways, we can choose 10 objects.

0 identical +10 distincts, number of ways =1×21C10

1 identical +9 distincts, number of ways =1×21C9

2 identicals +8 distincts, number of ways =1×21C8

. . . . . .

So, total number of ways in which we can choose 10 objects is

21C10+21C9+21C8++21C0=x (let)

21C11+21C12+21C13++21C21=x

On adding both Eqs. (i) and (ii), we get

[nCr=nCnr]

2x=21C0+21C1+21C2++21C10+21C11+21C12++21C212x=221x=220



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