Number of Arrangements of n Objects of m Types/Examples/6 people in 3 pairs

Example of Use of Number of Arrangements of $n$ Objects of $m$ Types
Let $N$ be the number of ways $6$ people can be partitioned into $3$ (unordered) pairs.

Then:
 * $N = 20$

Proof
Here we have an instance of Number of Arrangements of $3 p$ Objects into $3$ Equal Sized Subsets, such that $p = 2$.

Hence we have:
 * $N = \dfrac {\paren {3 \times 2}!} {\paren {2!}^3 \times 3!} = \dfrac {720} {8 \times 6} = \dfrac {720} {48} = 15$

Hence the result.