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.

  • $\begingroup$ Are the people distinguishable? $\endgroup$ – Bram28 Feb 3 '18 at 22:51
  • $\begingroup$ I think you're limiting too much in part 2. Can person 1 get orange and black and person 2 get orange and blue? $\endgroup$ – stuart stevenson Feb 3 '18 at 22:53
  • $\begingroup$ Part 1 looks good. $\endgroup$ – Joffan Feb 3 '18 at 23:02
  • $\begingroup$ In Part 1): If we give Ann blue and red and Bob white and black, is that different from giving Ann white and black and Bob blue and red? If the answer is "yes," then your answer is correct. If the answer is "no," then you shouldn't multiply by $k!$ $\endgroup$ – saulspatz Feb 3 '18 at 23:10
  • $\begingroup$ As to your question in part 2, you know that at the end, you have $n-2k$ colors remaining. How many were there before you chose the last pair of colors? $\endgroup$ – saulspatz Feb 3 '18 at 23:14

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.

  • $\begingroup$ There aren't any duplicates. Each person gets two different colors, and no two persons get the same color, so each color is used at most once. $\endgroup$ – saulspatz Feb 3 '18 at 23:12
  • $\begingroup$ For clarity - by "duplicates" I meant duplicate counting of any given colour combination - for example, red then yellow is the same as yellow then red. $\endgroup$ – Joffan Feb 4 '18 at 6:55
  • $\begingroup$ Ah, I see what you mean now. $\endgroup$ – saulspatz Feb 4 '18 at 7:00

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