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I am working on a problem where it says that we have 100 people and we are going to make pairs of 2 out of those people. how many ways are there to make such pairs?

per my understanding, we can use the permutation (100 2) to make such subsets. but this does not align or seems faulty.

For example, I have a set of 6 people {A,B,C,D} - using a Permutation I can get P(4,2) = 12

by permutation formula, I could have 12 pairs, but I could get only 3 such sets out of 4 people making pairs.

  1. {a,b},{c,d}
  2. {a,c},{d,b}
  3. {a,d},{c,b}

so, what should be the right way to solve this problem?

Atticus Stonestrom
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Zaniya Rizvi
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2 Answers2

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Permutations care about order, i.e. $(123)$ is a different permutation from $(132)$.

However when choosing pairs as described in your question, you neither care about the order of people inside each pair [$(12) \cong (21)]$, nor the order of the pairs themselves [$(12)(34) \cong (34)(12)$]. So we must adjust the permutation calculation accordingly.

There are $100!$ unique ways to line up the 100 children in order: $1\,2|3\,4|5\,6|...|99\, 100$ would be one such ordering. To take into account that the order of people inside each pair doesn't matter, divide by $2^{50}$ (once for each pair). To take into account that the order of the pairs doesn't matter, divide by $50!$.

Hence the answer is $\frac{100!}{2^{50}50!}$

packetpacket
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@lulu has given a reference where the problem has been solved in a number of ways,$\;$ some of them quite clever.

I am adding an almost mechanical approach, making use of one interpretation of the multinomial coefficient as selecting people for labeled groups.

So were the groups labeled, the answer for $2n$ people would be $\dfrac{(2n)!}{2!2!2!... to\;n\;terms} = \dfrac{(2n)!}{2^n}$

But since the groups are not labeled (either by distinct name or size), we divide by $n!\,,\;$ thus $\;\boxed{\dfrac{(2n)!}{2^nn!}}$

true blue anil
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