Question

Find the number of ways in which $8$ non-identical apples can be distributed among $3$ boys such that every boy should get at least $1$ apple & at most $4$ apples.

• A. $1320$
• B. $4620$
• C. $720$
• D. $1000$

Answer 1

Answer: (B)

$4620$

The apples are non-identical which we need to take care. Now the only possible way is

$\begin{array}{ccc}Boy1& Boy2& Boy3\end{array}$

$\begin{array}{ccc}4& 1& 3\\ 3& 2& 2\\ 4& 2& 2\end{array}$

Applying permutation combination above scenario and among boys we get

$=3![{{}^{8}C}_4\times {{}^{4}C}_1\times {{}^{3}C}_3]+3[{{}^{8}C}_3\times {{}^{5}C}_2\times {{}^{2}C}_2]+3[{{}^{8}C}_4\times {{}^{4}C}_2\times {{}^{2}C}_2]$

$=4620.$

Hence, the answer is $4620.$