How Many Ways to distribute six different books among 13 children if no child gets more than one book?
So I think the issue right now is I may have thought too deeply about the question.
Initially I thought the solution was C(13,6), but then I began to wonder what exactly the question is asking. Are there 6 different types of books and each child will get one? which would mean a solution like $6^{13}$.
Then I also thought, "what about the children who don't get books, do I have to account for all the different ways that can occur too? So that would be something of the sort $C(13,6) X 7!$
I know I am overthinking, but how far am I going?
Best Answer
You should choose the six children that will receive the books, i.e., $\binom{13}{6}$, and then you should multiple it by the number of permutations of the six different books among the six children, i.e., $6!$, thus $\binom{13}{6}6!$ .