We have $21$ different books and we have to distribute them to $3$ persons: $A$, $B$ and $C$. $A$ and $B$ together must have twice the number of books that $C$ has. What is the sum of the total ways to do it?
So:
\begin{align*}
A+B &= 2C,\\
A+B+C &= 21,\\
2C+C &= 21,\\
C &= 7,\\
A+B& = 14.
\end{align*}
I have been thinking the answer must be:
$$S = 2\cdot(\frac{21!}{1!\cdot13!\cdot7!} + \frac{21!}{2!\cdot12!\cdot7!} + \dots + \frac{21!}{7!\cdot7!\cdot7!}),$$
because $A$ can have $1$ book and $B$ can have $13$ books, $A$ can have $2$ books and $B$ can have $12$ books, …, $A$ can have $7$ books and $B$ can have $7$ books.
Then I multiply with $2$ because $A$ can have $13$ books and $B$ can have $1$ book, etc. So by multiplying with $2$ I make sure that I have all the distributions because $A$ and $B$ are different persons.
Is this solution correct? Any more elegant solution?
Best Answer
There are $21$ different books. $C$ gets $7$ books and rest $14$ books are divided between $A$ and $B$.
First, number of ways to choose $7$ books for $C = {21 \choose 7}$
Then, for each of the rest $14$ books, there are two choices - it either goes to $A$ or to $B$. So number of ways to distribute books to $A$ and $B$ is $ \ 2^{14}$. This includes cases where either $A$ or $B$ does not receive any book and all $14$ books go to the other.
So total number of ways $ = \displaystyle \small {21 \choose 7} \cdot 2^{14}$