1. Evaluate
(i) $8!$
$8!=1*2*3*4*5*6*7*8=40320$

(ii) $4!-3!$
$4!=1*2*3*4=24$
$3!=1*2*3=6$
$4!-3!=24-6=18$
2. Is $3!+4!=7!$?
$3!=1*2*3=6$
$4!=1*2*3*4=24$
$7!=1*2*3*4*5*6*7=5040$
$3!+4!=6+24=30$
∴ $3!+4!\neq 7!$
3. Compute $\frac{8!}{6!\times 2!}$
$\frac{8!}{6!\times 2!}$ $=\frac{8\times 7\times 6!}{6!\times 2!}$
= $\frac{8\times 7}{2!}$
= $\frac{8\times 7}{2}$
= $28$
4. If $\frac{1}{6!}+\frac{1}{7!}=\frac{x}{8!}$, find x.
$\frac{1}{6!}+\frac{1}{7!}=\frac{x}{8!}$
$\frac{1}{6!}+\frac{1}{7\times 6!}=\frac{x}{8\times 7\times 6!}$
$\frac{1}{6!}[1+\frac{1}{7}]=\frac{1}{6!}[\frac{x}{8\times 7}]$
$1+\frac{1}{7}=\frac{x}{8\times 7}$
$\frac{8}{7}=\frac{x}{8\times 7}$
$8=\frac{x}{8}$
$x=64$
5. Evaluate $\frac{n!}{(n-r)!}$, when
(i) $n=6, r=2$
$\frac{n!}{(n-r)!}$ $=\frac{6!}{(6-2)!}$
=$\frac{6!}{4!}$
=$\frac{6\times 5\times 4!}{4!}$
=$6\times 5=30$

(ii) $n=9, r=5$
$\frac{n!}{(n-r)!}$ $=\frac{9!}{(9-5)!}$
=$\frac{9!}{4!}$
=$\frac{9\times 8\times 7\times 6\times 5\times 4!}{4!}$
=$9\times 8\times 7\times 6\times 5=15120$