This question comes from a book.
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Combinatorics, dividing 11 distinct books among 3 people?
Suppose we have 15 different books, And we want these books to be split between 3 people.
The particular set up in the book is as follows:
"Ten different books are to be given to Daniel, Philip, Paul and John, who will get in the order 3, 3, 2, 2 books respectively.
Q1.) In how many ways can this be done?
Paul and John Scream "no fair" so they draw lots to decide which two get three and which two get two.
Q2.) How many ways are there now for a distribution.
Marilda and Corinna also want a chance and so it is decided that the six kids should draw lots to determine which two get 3, which two get 2 and which two get none.
Q3.) Now how many ways are there?"
This question come from a book and I understand the solution for the first two questions.
Answer 1 $\frac{10!}{3!3!2!2!}$
Answer 2 $\frac{10!}{3!3!2!2!} * \frac{4!}{2!2!}$
Answer 3 $\frac{10!}{3!3!2!2!} * \frac{6!}{2!2!2!2}$
What confuses me about answer 3 is the last 2 in the denominator of $\frac{6!}{2!2!2!2}$.
Why is it $\frac{6!}{2!2!2!2}$ rather than just $\frac{6!}{2!2!2!}$ which would be analogous to answer 2?
Your thinking is the correct, the final 2 in the denominator is likely a typo, since we can once again divide the six people into three categories with respectively 3,2 and 0 books.
Do however note that the claim you make in the final sentence is false since $$\frac{6!}{2!2!2!} = 15\cdot \frac{4!}{2!2!} \neq \frac{4!}{2!2!}$$