Answer
$9900$
Work Step by Step
See review summary:
b. Factorial Notation
$n!=n(n-1)(n-2)\cdots(3)(2)(1) $ and $0!=1 $(by definition)
( It follows that $n!=n(n-1)!$ )
c. Permutations Formula$ \displaystyle \quad {}_{n}P_{r}=\frac{n!}{(n-r)!}$
---
${}_{10}P_{6}=\displaystyle \frac{100!}{(100-2)!}=\frac{100!}{98!}\qquad $... apply $n!=n(n-1)!$
=$\displaystyle \frac{100\times 99!}{98!}=\frac{100\times 99\times 98!}{98!}$
... reduce the fraction (divide with $\displaystyle \frac{98!}{98!}$)
$=100\times 99$
= $9900$
