Question

The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is:
(a) 5
(b) 21
(c)${}^{8}{{C}_{3}}$
(d)${{3}^{8}}$

Answer 1

Hint: Let us assume the number of balls contained in the three boxes is $x,y\And z$ respectively. Now, the sum of these three balls must be 8 because it is given that we have to distribute 8 identical balls in 3 distinct boxes so add x, y and z then equate this summation to 8. Also, there is a condition given in the above problem that none of the boxes is empty so the minimum value that x, y and z can take is 1. Now, using this condition we will solve the above problem.

Complete step by step solution:
We have given 8 identical balls and 3 distinct boxes and we are asked to find the number of ways to distribute 8 identical balls in 3 distinct boxes.
Let us suppose that the three distinct boxes are A, B and C and these three boxes contain x, y and z number of balls respectively. Then the sum of x, y and z will be 8.
$x+y+z=8$ ……. (1)
But there is a condition which states that none of the boxes is empty so there is a condition which is imposing on x, y and z are:
$\begin{align}
  & x\ge 1, \\
 & y\ge 1, \\
 & z\ge 1 \\
\end{align}$
Now, we know the condition that if:
$a+b+c=n$
In this condition we have r identical things which we have to distribute in 3 distinct boxes (a, b and c) and (a, b, c) (n) can take values from 0 to n and the number of ways are:
${}^{n+r-1}{{C}_{r}}$
In the above formula, “r” is the identical items and “n” are the distinct items whereas in the above solution (x, y, z) can take value from 1 so to apply the above formula we have to make x, y and z non – negative by writing $\left( l+1 \right),\left( m+1 \right),\left( n+1 \right)$ respectively in place of x, y and z in eq. (1) we get,
$\begin{align}
  & l+1+m+1+n+1=8 \\
 & \Rightarrow l+m+n=8-3=5 \\
 & \Rightarrow l+m+n=5.......(2) \\
\end{align}$
Now, using the formula ${}^{n+r-1}{{C}_{r}}$ in the above equation by putting the value of r as 5 and n as 3 we get,
$\begin{align}
  & {}^{n+r-1}{{C}_{r}} \\
 & ={}^{3+5-1}{{C}_{5}} \\
 & ={}^{8-1}{{C}_{5}}={}^{7}{{C}_{5}} \\
 & ={}^{7}{{C}_{2}} \\
\end{align}$
Solving the above combinatorial expression we get,
$\begin{align}
  & \dfrac{7!}{2!\left( 7-2 \right)!} \\
 & =\dfrac{7!}{2!\left( 5 \right)!} \\
\end{align}$
We can write $7!=7.6.5!$ in the above expression and we get,
$\dfrac{7.6.5!}{2!\left( 5! \right)}$
In the above expression, $5!$ will be cancelled out from the numerator and the denominator and we get,
$\begin{align}
  & \Rightarrow \dfrac{7.6}{2.1} \\
 & =\dfrac{42}{2} \\
 & =21 \\
\end{align}$

So, the correct answer is “Option b”.

Note: The mistake that could be possible in the above problem is that you might have used the formula ${}^{n+r-1}{{C}_{r}}$ in eq. (1) of the above solution which would be wrong to use so make sure you will not miss that point of converting eq. (1) to eq. (2).