Question

Five different balls are to be placed in 5 different boxes randomly. If each box can hold any number of balls, the number of ways in which exactly 2 boxes remain empty is?
A.1200
B.1600
C.1400
D.1500

Answer 1

Hint: When $ n $ objects are to be placed in $ r $ seats randomly, then this can be done in $ {n^r} $ ways. Use the formula $ {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $ in this question to reduce the complication of this question. Also, the number of ways of arranging $ n $ things is in $ n! $ ways, this formula will help to find out the number of ways to arrange things.

Complete step-by-step answer:
We know that, when $ n $ objects are to be placed in $ r $ seats randomly, then this can be done in $ {n^r} $ ways.
In this question, 5 different balls are to be placed in 5 different boxes randomly. We can compare this situation with the situation where $ n $ objects are to be placed in $ r $ seats randomly which can be done in $ {n^r} $ ways. Substitute $ n = 5,{\rm{ }}r = 5 $ in $ {n^r} $ to find the required value.
5 different balls are to be placed in 5 different boxes, then this can be done in $ {5^5} $ ways.
In the condition, as given in the question, 2 out of 5 boxes are too empty.
In order to select $ r $ objects out of $ n $ objects, it can be selected in $ {}^n{C_r} $ .
2 empty boxes can be selected in $ {}^5{C_2} $ ways.
So, 5 balls can now be placed in $ 5 - 2 = 3 $ boxes.
There can be 3,1,1 or 2,2,1 balls in the boxes.
We know that the number of ways of arranging $ n $ things is in $ n! $ ways.
So, 5 different balls can be arranged as 3,1,1 in $ \dfrac{{5!}}{{3! \cdot 2!}} $ where 3, 1 and 1 balls are present in different boxes but 1 is repeated twice so we need to divide by $ 2! $ .
Similarly, 5 different balls can be arranged as 2,2,1 in $ \dfrac{{5!}}{{2! \cdot 2!1!}} $ where 2, 2 and 1 balls are present in different boxes but 2 is repeated twice so we need to divide by $ 2! $ again.
The 3,1,1 or 2,2,1 way can be added to each other as they are 2 different cases.
Also, the 3 boxes are to be arranged in $ 3! $ ways.
So, as all the steps are necessary at a time, so all the factors should be multiplied to each other in order to reach the solution.
The number of ways in which exactly 2 boxes remain empty is $ 3!\left( {\dfrac{{5!}}{{3! \cdot 1! \cdot 1!}} + \dfrac{{5!}}{{2! \cdot 2! \cdot 1!}}} \right) $
 $
\Rightarrow 3!\left( {\dfrac{{5!}}{{3! \cdot 2!}} + \dfrac{{5!}}{{2! \cdot 2! \cdot 2!}}} \right) = 6\left( {\dfrac{{120}}{{12}} + \dfrac{{120}}{8}} \right)\\
 = 6\left( {10 + 15} \right)\\
 = 6 \times 25\\
 = 150
 $

The number of ways in which exactly 2 boxes remain empty is $ {}^5{C_2} \cdot 150 = 1500 $ .

So, the correct answer is “Option D”.

Note: Students must avoid mistakes while using the formula for calculating the number of ways when $ n $ objects are to be placed in $ r $ seats randomly. Instead of taking the number of ways as $ {n^r} $, students can take it mistakenly as $ {r^n} $ . Also, the language of the question should be understood in a clear manner so that conditions are well expressed in terms of mathematical symbols.