A shopkeeper distributes 12 identical sweets among 4 distinct children so that each child gets at least 1 and at most 5 sweets. The number of ways to…

Let $a_i$ be the number received by child $i$ and put $b_i=a_i-1$, so $b_1+b_2+b_3+b_4=8$ with $0\le b_i\le4$. Without the upper bound there are $\binom{11}{3}=165$ solutions. A violation means some $b_i\ge5$; setting $b_i=5+c_i$ leaves $c$-sum $3$, giving $\binom63=20$ solutions per child, and $4\cdot20=80$ in total. Two children cannot both violate since $5+5>8$, so no overlap correction is needed. The answer is $165-80=85$. The option $165$ ignores the upper limit, and $125$ arises if the minimum of $1$ is dropped but the maximum of $5$ is kept (counting solutions with $0\le a_i\le5$).