## Precalculus (6th Edition) Blitzer

The required solution is ${}_{n}{{C}_{r}}=\frac{n!}{r!\left( n-r \right)!}$
We know that the above problem uses combinations as the order of the selection does not matter. The formula is given by: ${}_{n}{{C}_{r}}=\frac{n!}{r!\left( n-r \right)!}$