How many ways to group $5$ into two pairs and a single? I get $30$, but the answer is $15$. How can I tell that I over-counted?

Question

Suppose I have 5 people and I want to group them into two pairs and a single person. How many ways can I do this?

Here is my naive thinking using the binomial coefficient $$\binom{5}{2}\binom{3}{2}\binom{1}{1} = \frac{5!}{2^2} = 30$$ However, I've over-counted: I know there should be 15 combinations, but I have 30.

How can I justify dividing by 2?

Answer 1

Observe that $(AB)(CD)E$ and $(CD)(AB)E$ are actually equal results.

You counted them twice.

So the correct outcome is:$$\frac1{2!}\binom{5}{2}\binom{3}{2}\binom{1}{1} = 15$$

Answer 2

You can choose the single person in $5$ ways. The eldest of the remaining $4$ persons can choose his partner in $3$ ways, and then everything is decided. Makes $5\cdot3=15$ possibilities.