Splitting books between 3 people such that two will have twice the number as the other There are 21 books and 3 people (a, b and c). How many ways can the books be distributed such that two people together will have twice as many books as the third?
My method is $\binom{3}{2}\binom{21}{14}$ which is simply choosing two people to have the "twice as many books" and the "twice as many books" is fulfilled by choosing 14 out of the 21 total
I am quite confident in the answer but this seems way too simple and I was provided quite a lot of space for working out for this question so it seems I may be missing something. If that is the case, please provide hints and blur the answer
 A: If the books are distinguishable and one person receives exactly seven books, then you still need to count the number of ways to distribute the 14 books among the two people. You had the correct start:
$$\dbinom{3}{2}\dbinom{21}{14}$$
This chooses two of the three people to receive 14 books (the last person receives seven by default). It then chooses 14 of the 21 distinguishable books (the last person receives the remaining seven books by default). Now, among the 14 books, you need to distribute them to the remaining people. How many ways are there to distribute 14 books to 2 people where every book gets distributed? You can look at it like the first book can go to the first person or the second person. The second book can go to the first person or the second person. Etc.
Finally, you need to subtract any overcounting. This counting method I described counts situations where you distribute seven books to all three people three times each.

 $$\dbinom{3}{2}\dbinom{21}{14}2^{14} - 2\dfrac{21!}{(7!)^3}$$

