Numbers of ways to give 2 differently colored balloons

Question

We have unlimited supply of balloons in n colors. Calculate the number of ways to give 2 differently colored balloons to each of k different people if:

a. no 2 people can get the same pair of colors

b. no 2 people can get the same color

Answer 1

a.

There are $n \choose 2$ color pairs to choose from. Now choose the pairs to give to the people one by one and we count the ways as the product

$${n \choose 2} \times \left({n \choose 2}-1\right) \times \dots \times \left({n \choose 2}-(k-1)\right). $$

Note: this is valid when $k\leq {n \choose 2}$. Otherwise there are $0$ ways because there isn't enough color pairs to give (well, then $0={n \choose 2}-{n \choose 2}$ appears in the above product so I guess it is valid also in that case).

b.

Now there must be at least $2k$ colors. That is, $n \geq 2k$. Let's do the assignation of colors like this:

• Order the people in line (note: this can be done arbitrarily, this is the "fixing step")
• Choose the $2k$ colors to use. This can be done ${n} \choose {2k}$ ways.
• Choose an order for the colors (the colors are then given to the people like this: two from the start to the first, the next two to the second and so on). The ordering can be done $(2k)!$ ways but two colors are given to each person and it doesn't matter in which order those two colors are given, so we must divide by $(2!)^k = 2^k$.

The total number of ways is thus

$$\frac{{{n} \choose {2k}} (2k)!}{2^k} = \frac{{\frac{n!}{(2k)!(n-2k)!}} (2k)!}{2^k} = \frac{n!}{2^k(n-2k)!}. $$