Fraction Of Factorials

Note that $n!=n \times (n-1) \times (n-2) \times ... \times 1$ where $n>0$

$\frac {9!+8!}{8!}$

$=\frac {9(8!)+8!}{8!}$

$=\frac {8!(9+1)}{8!}$

$=9+1$

$=\boxed{10}$

$\Rightarrow \dfrac{9!+8!}{8!}$

$\Rightarrow \dfrac{8!(9+1)}{8!}=\boxed{10}$

Note that

• $n! = n \times (n-1) \times \dots \times 2 \times 1$
• $(n+1)! = (n+1) \times n!$

Now,

$\frac {8! + 9!}{8!}$

$=$ $\frac {9 \times (8!) +1 \times (8!)}{8!}$

$= \frac {(9+1) \times (8!)}{8!}$

$= \frac {10 \times (8!)}{8!}$

$= 10 \times \frac {8!}{8!}$

$= 10 \times 1$

$= \boxed{10}$

$\frac{9!+8!}{8!}=\frac{9!}{8!}+\frac{8!}{8!}=\frac{9!}{8!}+1=9+1=\boxed{10}$