Question

Number of ways in which 6 distinct objects can be kept into two identical boxes so that no box remains empty is

A) 31

B) 32

C) 63

D) 64

Answer 1

Here, we will use the concept of permutation. Firstly, we will find the number of ways 6 distinct objects can be kept in two identical boxes. Here we will be using the permutations to get this. Then we have to consider the number of ways any one of the boxes can be empty. Lastly we just need to subtract both the numbers to get our final answer.

Complete step by step solution:

Number of distinct objects $= 6$

Number of boxes $= 2$

Number of ways in which a single object can be kept into one of the two boxes $= {}^2{P_1} = \dfrac{{2!}}{{(2 - 1)!}}$

Simplifying the factorial, we get

$\Rightarrow$ Number of ways in which a single object can be kept into one of the two boxes $= \dfrac{{2!}}{{(1)!}} = 2$

In the similar way all the 6 boxes will be kept into two different boxes.

So, the total number of ways in which 6 distinct objects can be kept into two different boxes will be equal to the six times the number of ways in which a single object can be kept into one of the two boxes.

Therefore, number of ways in which 6 distinct objects can be kept into two different boxes $= 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^6 = 64$

But here we have two identical boxes.

Therefore the number of ways in which 6 distinct objects can be kept into two identical boxes $= \dfrac{64}{2} = 32$.

We have still not reached the answer. This is not the answer as per the question. Because in the question it is mentioned that, no boxes should be empty.

So here any one of the boxes can be empty. If these two boxes would be different then the number of ways any one box can be empty will be equal to 2.

But we have two identical boxes.

So the number of ways any one of the boxes can be empty will be 1.

In the question, it is said that no boxes should be empty. So we have to subtract 1.

Therefore the number of ways in which 6 distinct objects can be kept into two identical boxes = 32-1 = 31

So, option (A) is the correct option.

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Note: Here we have to note that we need to apply the concept of permutation, not the combination for the arrangement of the letters. A permutation is used to find the different ways in which a collection of items can be arranged in a certain sequence or order. Whereas combination is used to find the various ways in which objects from a set may be selected such that order of selecting an object doesn’t matter.