# Question 3f31e

May 16, 2017

Answer: $21$

#### Explanation:

Evaluate $\left(\begin{matrix}7 \\ 5\end{matrix}\right)$

This is a combination, in general form:
((n),(k))=(n!)/(k!(n-k)!)

So, for this problem, we have:
$\left(\begin{matrix}7 \\ 5\end{matrix}\right)$
=(7!)/(5!(7-5)!)
=(7!)/(5!(2!))

Note that we can re-write 7! as 7*6*5! to cancel with the 5! in the denominator:
=(7*6*5!)/(5!(2!))#
$= \frac{7 \cdot 6}{2}$
$= 7 \cdot 3$
$= 21$