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

\[

(A){\text{ }}{2^n} \\

(B){\text{ }}{2^{n - 1}} \\

(C){\text{ }}{2^n} - 1 \\

(D){\text{ }}{2^n} + 1 \\

\]

Last updated date: 23rd Mar 2023

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Answer

Verified

311.1k+ views

Hint:- Use the concept of combinations.

Let \[{S_1}\] be the set of n identical objects

And \[{S_2}\] be the set of the remaining n different objects.

So, now the number of ways to select r objects from \[{S_1}\] since all objects in \[{S_1}\] are identical will be 1.

Number of ways to select r objects from \[{S_2}\] will be \[{}^n{C_r}\].

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 \[{S_2}\] and 0 object from \[{S_1}\]

Number of ways \[ = {}^n{C_n}\]

Case 2:

N-1 objects from \[{S_2}\] and 1 object from \[{S_1}\]

Number of ways \[ = {}^n{C_{n - 1}}*1 = {}^n{C_{n - 1}}\]

Case 3:

N-2 objects from \[{S_2}\] and 2 objects from \[{S_1}\]

Number of ways \[ = {}^n{C_{n - 2}}*1 = {}^n{C_{n - 2}}\]

Case 4:

N-3 objects from \[{S_2}\] and 3 objects from \[{S_1}\]

Number of ways \[ = {}^n{C_{n - 3}}*1 = {}^n{C_{n - 3}}\]

.

.

.

.

.

Case n-1:

2 objects from \[{S_2}\] and n-2 objects from \[{S_1}\]

Number of ways \[ = {}^n{C_2}*1 = {}^n{C_2}\]

Case n:

1 object from \[{S_2}\] and n-1 objects from \[{S_1}\]

Number of ways \[ = {}^n{C_1}*1 = {}^n{C_1}\]

Case n+1:

0 object from \[{S_2}\] and n objects from \[{S_1}\]

Number of ways \[ = {}^n{C_0}*1 = {}^n{C_0}\]

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 \[ = {}^n{C_n} + {}^n{C_{n - 1}} + {}^n{C_{n - 2}} + {}^n{C_{n - 3}} + ..... + {}^n{C_2} + {}^n{C_1} + {}^n{C_0}\]

As we know that according to binomial theorem,

\[ \Rightarrow {}^n{C_n} + {}^n{C_{n - 1}} + {}^n{C_{n - 2}} + {}^n{C_{n - 3}} + ..... + {}^n{C_2} + {}^n{C_1} + {}^n{C_0} = {2^n}\]

So, total number of ways \[ = {2^n}\]

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.

Let \[{S_1}\] be the set of n identical objects

And \[{S_2}\] be the set of the remaining n different objects.

So, now the number of ways to select r objects from \[{S_1}\] since all objects in \[{S_1}\] are identical will be 1.

Number of ways to select r objects from \[{S_2}\] will be \[{}^n{C_r}\].

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 \[{S_2}\] and 0 object from \[{S_1}\]

Number of ways \[ = {}^n{C_n}\]

Case 2:

N-1 objects from \[{S_2}\] and 1 object from \[{S_1}\]

Number of ways \[ = {}^n{C_{n - 1}}*1 = {}^n{C_{n - 1}}\]

Case 3:

N-2 objects from \[{S_2}\] and 2 objects from \[{S_1}\]

Number of ways \[ = {}^n{C_{n - 2}}*1 = {}^n{C_{n - 2}}\]

Case 4:

N-3 objects from \[{S_2}\] and 3 objects from \[{S_1}\]

Number of ways \[ = {}^n{C_{n - 3}}*1 = {}^n{C_{n - 3}}\]

.

.

.

.

.

Case n-1:

2 objects from \[{S_2}\] and n-2 objects from \[{S_1}\]

Number of ways \[ = {}^n{C_2}*1 = {}^n{C_2}\]

Case n:

1 object from \[{S_2}\] and n-1 objects from \[{S_1}\]

Number of ways \[ = {}^n{C_1}*1 = {}^n{C_1}\]

Case n+1:

0 object from \[{S_2}\] and n objects from \[{S_1}\]

Number of ways \[ = {}^n{C_0}*1 = {}^n{C_0}\]

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 \[ = {}^n{C_n} + {}^n{C_{n - 1}} + {}^n{C_{n - 2}} + {}^n{C_{n - 3}} + ..... + {}^n{C_2} + {}^n{C_1} + {}^n{C_0}\]

As we know that according to binomial theorem,

\[ \Rightarrow {}^n{C_n} + {}^n{C_{n - 1}} + {}^n{C_{n - 2}} + {}^n{C_{n - 3}} + ..... + {}^n{C_2} + {}^n{C_1} + {}^n{C_0} = {2^n}\]

So, total number of ways \[ = {2^n}\]

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.

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