Number of ways to put $2r$ distinguishable balls in $r$ distinguishable boxes where exactly $r$ boxes contain exactly two balls

Question

Number of ways to put $2r$ distinguishable balls in $r$ distinguishable boxes where exactly $r$ boxes contain exactly two balls.

Actually, that's all.

Well, I can tell you my solution. \begin{eqnarray} \left(\begin{array}{c} 2 \\ 2 r \end{array}\right)\left(\begin{array}{c} 2 \\ 2 r-2 \end{array}\right)\left(\begin{array}{c} 2 \\ 2 r-4 \end{array}\right) \cdots\left(\begin{array}{l} 2 \\ 2 \end{array}\right) r ! \end{eqnarray}

But I want to get some analytical solution to this problem. In other words, I want some help to simplify this or another solution.

Answer 1

What you have written as $\displaystyle {2 \choose 2r}$ would normally be written as $\displaystyle {2r \choose 2}$ etc. Then use the definition $\displaystyle {n \choose r} = \dfrac{n!}{r!(n-r)!}$ on each factor and you will get lots of cancellation, leading to quite a nice final answer.

It might be worth adding that the easiest way of doing the problem is to note that there are $(2r)!$ ways of arranging the balls in a line; then just put the first two in the first box, the next two in the second box, ... and so on. Now you have counted each arrangement of balls in boxes $2^r$ times, because you have counted each pair in each box in both possible orders in the line. So altogether $\dfrac{(2r)!}{2^r}$ different arrangements.

Answer 2

Your answer is too large. Suppose we have two balls and one box. There is only one way to place both balls in the same box. Your answer gives $\binom{2}{2}2! = 2$. Multiplying by $r!$ is not necessary.

In what follows, I will use the notation $$\binom{n}{k} = \frac{n!}{k!(n - k)!}$$ for the number of $k$-element subsets of a set with $n$ elements.

Line the boxes up from left to right. There are $\binom{2r}{2}$ ways to select which two balls will be placed in the first box, $\binom{2r - 2}{2}$ ways to select which two of the remaining balls will be placed in the second box, $\binom{2r - 4}{2}$ ways to select which two of the remaining balls will be placed in the third box, and so forth until we must place the final two balls in the last box. Hence, there are $$\binom{2r}{2}\binom{2r - 2}{2}\binom{2r - 4}{2} \cdots \binom{4}{2}\binom{2}{2}$$ ways to distribute two distinct balls from a set of $2r$ distinct balls to each of $r$ distinct boxes.

This expression simplifies nicely: \begin{align*} \binom{2r}{2}\binom{2r - 2}{2}\binom{2r - 4}{2} \cdots \binom{4}{2}\binom{2}{2} & = \frac{(2r)!}{2!(2r - 2)!} \cdot \frac{(2r - 2)!}{2!(2r - 4)!} \cdot \frac{(2r - 4)!}{2!(2r - 6)!} \cdots \frac{4!}{2!2!} \cdot \frac{2!}{2!0!}\\ & = \frac{(2r)!}{(2!)^n0!}\\ & = \frac{(2r)!}{2^n} \end{align*}