Question: How many ways can you give seven same candy to three children? (Every child should be given at least $1$ candy)
My solution:
$$\begin{cases}m+n+k=7 \\1≤m≤5 \\1≤n≤5 \\ 1≤k≤5 \end{cases}$$
All possible solutions are:
$115,124,133, 142, 151, 214, 223,231, 241,313,321, 331, 412, 421,511$
So, we have $5+4+3+2+1=15$ different ways.
Is my solution correct?
But the answer I found is not in the possible answers. I can't see where I made a mistake.