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Q.
The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is
Permutations and Combinations
Solution:
We know that the number of ways of distributing n identical items among r persons, when each one of them receives at least one item is ${^{n-1}C_{r-1}}$
$\therefore $ The required number of ways
$= ^{8-1}C_{3-1}= ^{7}C_{2} = \frac{7!}{2!5!}= \frac{7 \times6}{2 \times1} = 21 $