In how many ways can 4 distinct chocolates be distributed among three children such that exactly one of the children gets no chocolate?
Let's choose two children in $\binom{3}{2}$ ways whom we have to distribute. Total number of ways to distribute four chocolates among them is hence $\binom{3}{2}2^4$. However , in the total ways to distribute 4 chocolates , which is $2^4$ , I've also counted the cases when one of those two children gets no chocolate (which is same as the case when only one child gets all the chocolates). To compensate for that, I subtracted $\binom{3}{1}$ from the total ways. Hence, the correct final answer should be :
$$\binom{3}{2}2^4-\binom{3}{1}=45$$
However, according to answer key, the answer is $42$. Where did I go wrong ?
Best Answer
Let's correct your approach.
There are $\binom{3}{2}$ ways to select the two children will receive chocolates. If there were no restrictions, there would be $2^4$ ways to distribute the chocolates to the two selected students. However, those distributions include two distributions in which only one of the two selected students receives chocolates. Hence, the number of ways to distribute four distinct chocolates to three children such that exactly one of the children receives no chocolates is $$\binom{3}{2}(2^4 - 2)$$