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

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- $C(n,r)=\large\frac{n!}{r!(n-r)!}$
- $n!=n(n-1)(n-2).....(3)(2)(1)$

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.

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