Dividing people into pairs, permutations and combinations

Question

How do we calculate the no of ways of dividing 100 people into 50 pairs?

Answer 1

Think of finding a first pair, then a second, and so on. Let $A$ be the number of ways of doing this. Since we're (presumably) interested in the number of ways of pairing the people with the pairs unordered, the answer will be $A/50!$ (there are $50!$ ways to line up $50$ given pairs; each of these is counted in the preceding).

There are $100\choose2$ ways of selecting the first pair. There are $98\choose2$ ways of selecting the second. There are $96\choose 2$ ways of selecting the third. ... By the multiplication principle, $A$ is the product of the these quantities. After simplifying, one obtains $A={100!\over 2^{50}}$.

So there are $100!\over 50!\cdot 2^{50}$ ways of dividing $100$ people into $50$ pairs.

Answer 2

Order the people by height. The smallest person has $99$ partner choices.

For every choice made by the smallest person, the smallest unchosen person has $97$ choices.

For every one of the first $2$ choices, the smallest unchosen person has $95$ choices.

And so on.

The total number of choices is $$(99)(97)(95)\cdots (5)(3)(1).$$ We can modify the shape of the answer by multiplying top and "bottom" by $(100)(98)(96)\cdots (4)(2)$, that is, by $2^{50}50!$. The top becomes $100!$, so an alternate expression is $$\frac{100!}{2^{50}50!}.$$