How many ways are there to give 21 different books to 3 people, A, B, C, so that A's + B's = 2C's books? Only thing I found is that A and B will take 14 books, while C has to get 7. Any thoughts?
Thanks in advance.
Edit:
So the question is not answered, we are looking for ways to give the books to the three people with the above conditions, though I might have come up with a solution, which I'd be glad to discuss about.
Solution (?):
First of all, we give 7 books to C, but how many ways are there? C(n,r) = C(21, 7). Then the remaining 14 books have to be given to A and B, but B is actually not having any impact, you can give A the 14 with $2^n$ -1 ways, n = 14. B will either way take the remaining books A didn't take.
Multiplying those 2 numbers we take the required ways.
Any thoughts?
If A gets k books out of a selection of 21, then B must get the rest of the 14 books (14-k) from the rest of the selection (21-k). Determining the books for A and B also determines the books for C.
$\displaystyle\sum_{k=0}^{14}\binom{21}{k}\binom{21-k}{14-k} = 1,905,131,520$
Hint:
Obviously you can take a book from A and give it to B (or the other way round) while still satisfying the conditions.
Write the two conditions as $$A+B+C=21$$ $$A+B=2C$$
You have two equations and three unknowns, so there is not a single unique solution, but in this example you can find $C$ and $A+B$ and it is not difficult to do so.  You are already most of the way there.