Solveeit Logo
Login

Question

Question: The number of selections of n objects from 2n objects of which n are identical and the rest are diff...

The number of selections of n objects from 2n objects of which n are identical and the rest are different is:

(A) 2n (B) 2n−1 (C) 2n−1 (D) 2n+1  (A){\text{ }}{2^n} \\\ (B){\text{ }}{2^{n - 1}} \\\ (C){\text{ }}{2^n} - 1 \\\ (D){\text{ }}{2^n} + 1 \\\
Explanation

Solution

Hint:- Use the concept of combinations.

Let S1{S_1}​ be the set of n identical objects
And S2{S_2}​ be the set of the remaining n different objects.
So, now the number of ways to select r objects from S1{S_1} since all objects in S1{S_1} are identical will be 1.
Number of ways to select r objects from S2{S_2} will be nCr{}^n{C_r}.
So, as we know that there will be many cases to select n objects from 2n objects.
And, sum of number of ways of all n+1 case will be equal to total number of ways to select n objects
from 2n objects.
Case 1:
N objects from S2{S_2} and 0 object from S1{S_1}
Number of ways =nCn = {}^n{C_n}
Case 2:
N-1 objects from S2{S_2} and 1 object from S1{S_1}
Number of ways =nCn−1∗1=nCn−1 = {}^n{C_{n - 1}}*1 = {}^n{C_{n - 1}}
Case 3:
N-2 objects from S2{S_2} and 2 objects from S1{S_1}
Number of ways =nCn−2∗1=nCn−2 = {}^n{C_{n - 2}}*1 = {}^n{C_{n - 2}}
Case 4:
N-3 objects from S2{S_2} and 3 objects from S1{S_1}
Number of ways =nCn−3∗1=nCn−3 = {}^n{C_{n - 3}}*1 = {}^n{C_{n - 3}}
.
.
.
.
.
Case n-1:
2 objects from S2{S_2} and n-2 objects from S1{S_1}
Number of ways =nC2∗1=nC2 = {}^n{C_2}*1 = {}^n{C_2}
Case n:
1 object from S2{S_2} and n-1 objects from S1{S_1}
Number of ways =nC1∗1=nC1 = {}^n{C_1}*1 = {}^n{C_1}
Case n+1:
0 object from S2{S_2} and n objects from S1{S_1}
Number of ways =nC0∗1=nC0 = {}^n{C_0}*1 = {}^n{C_0}
So, now total number of selections of n objects from 2n objects out of which n are identical
Will be the sum of all cases.
So, total number of ways =nCn+nCn−1+nCn−2+nCn−3+.....+nC2+nC1+nC0 = {}^n{C_n} + {}^n{C_{n - 1}} + {}^n{C_{n - 2}} + {}^n{C_{n - 3}} + ..... + {}^n{C_2} + {}^n{C_1} + {}^n{C_0}
As we know that according to binomial theorem,
⇒nCn+nCn−1+nCn−2+nCn−3+.....+nC2+nC1+nC0=2n\Rightarrow {}^n{C_n} + {}^n{C_{n - 1}} + {}^n{C_{n - 2}} + {}^n{C_{n - 3}} + ..... + {}^n{C_2} + {}^n{C_1} + {}^n{C_0} = {2^n}
So, total number of ways =2n = {2^n}
Hence the correct option will be A.

Note:- Whenever we came up with this type of problem then easiest and efficient way is to
Find different cases for selections of objects and then the total number of selections will be the sum of selections of all the cases.