Prompt: We have unlimited number of balloons in n colors. How many ways are there to give 2 differently colored balloons to each of k people if:
- no 2 people can get the same pair of colors.
- no 2 people get same color.
For part 1,
First we choose a pair of 2 distinct balloons = $n\choose2$
Then, we find he number of ways to choose these pairs = ${{n\choose 2}\choose{k}}$
The ways to give one of k different pairs to each of k different people = k!
Final answer : $${{{n}\choose{2}}\choose{k}}k!$$
For part 2,
Since no 2 people get the same color, we have to give each person a pair of different colors
So first person gets $n\choose2$ balloons
Second person gets $(n-2)\choose2$
Third person gets $(n-4)\choose2$
... and so on till the last balloon, but I'm not sure how to find the last person. Also I'm not sure about the method I used.
As an expression for part 2, assume the people are arbitrarily arranged, choose $2k$ colours and order, adjusting for duplicate cases per person:
$$\binom {n}{2k} \frac {2k!}{2^k} = \frac{n!}{(n-2k)!2^k} $$
In fact this is also what you get from continuing your analysis.
$$\require{cancel}\begin{align} \binom {n}{2}\cdot \binom {n-2}{2}&\cdot \binom {n-4}{2}\cdots \binom {n-2k+2}{2} \\ &= \frac{n!}{(n{-}2)!2!}\cdot \frac{(n{-}2)!}{(n{-}4)!2!}\cdot \frac{(n{-}4)!}{(n{-}6)!2!}\cdots \frac{(n{-}2k{+}2)!}{(n{-}2k)!2!} \\ &= \frac{n!}{\cancel{(n{-}2)!}2!}\cdot \frac{\cancel{(n{-}2)!}}{\cancel{(n{-}4)!}2!}\cdot \frac{\cancel{(n{-}4)!}}{\cancel{(n{-}6)!}2!}\cdots \frac{\cancel{(n{-}2k{+}2)!}}{(n{-}2k)!2!} \\ &=\frac{n!}{(n-2k)!2^k} \end{align}$$ as before.