## Precalculus (6th Edition) Blitzer

The required solution is $210$
We know that the representation $_{n}{{C}_{r}}$ counts the number of assortments of n items taken r at a time. And the number of assortment of n items taken r at a time can be evaluated as: $_{n}{{C}_{r}}=\frac{n!}{\left( n-r \right)!r!}$ And the provided expression is $_{10}{{C}_{6}}$. Here, $n=10,r=6$. Put the value of n, r in the above formula. Then: \begin{align} & _{10}{{C}_{6}}=\frac{10!}{\left( 10-6 \right)!6!} \\ & =\frac{10!}{4!6!} \\ & =\frac{10\cdot 9\cdot 8\cdot 7\cdot 6!}{4\cdot 3\cdot 2\cdot 1\cdot 6!} \end{align} And simplify further, \begin{align} & \frac{10\cdot 9\cdot 8\cdot 7\cdot 6!}{4\cdot 3\cdot 2\cdot 1\cdot 6!}=\frac{10\cdot 9\cdot 8\cdot 7}{4\cdot 3\cdot 2\cdot 1} \\ & =\frac{5040}{24} \\ & =210 \end{align} Hence, $_{10}{{C}_{6}}=210$