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:

$N = 15$

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.

$\blacksquare$

Sources

• 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text I$. Algebra: Permutations and Combinations: Exercises $\text I$: $3$