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# The number of ways of distributing 8 identical balls in 3 distinct boxes,so that none of the boxes is empty is

$\begin{array}{1 1}(A)\;5\\(B)\;21\\(C)\;3^8\\(D)\;8C_3\end{array}$

Toolbox:
• $C(n,r)=\large\frac{n!}{r!(n-r)!}$
• $n!=n(n-1)(n-2).....(3)(2)(1)$
The total no of ways of dividing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is
$(8-1)C_{3-1}=7C_2$
$\Rightarrow \large\frac{7!}{2!5!}$
$\Rightarrow \large\frac{7\times 6\times 5!}{2\times 5!}$
$\Rightarrow 21$
Hence (B) is the correct answer.