# How do you simplify (21!)/(17!4!)?

Oct 30, 2016

$5985$

#### Explanation:

(21!)/(17! xx 4!)=(17! xx 18 xx 19 xx 20 xx 21)/(17! xx4!)

may be simplified as

$= \frac{18 \times 19 \times 20 \times 21}{1 \times 2 \times 3 \times 4}$

and this may be further simplified as

$= \frac{9 \times 19 \times 5 \times 7}{1 \times 1 \times 1 \times 1}$

$= 9 \times 19 \times 5 \times 7$

$= 5985$

Oct 30, 2016

Write out the definitions of each factorial and you should get $5985$.

$\implies \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdots 17 \cdot 18 \cdot 19 \cdot 20 \cdot 21}{\left(1 \cdot 2 \cdot 3 \cdot 4 \cdots 15 \cdot 16 \cdot 17\right) \left(1 \cdot 2 \cdot 3 \cdot 4\right)}$

The portion of 21! that is before $18 \cdot 19 \cdot 20 \cdot 21$ cancels out.

$= \frac{18 \cdot 19 \cdot 20 \cdot 21}{1 \cdot 2 \cdot 3 \cdot 4}$

$= \frac{\left(19 - 1\right) \cdot 19 \cdot 20 \left(20 + 1\right)}{24}$

$= \frac{\left(361 - 19\right) \cdot \left(400 + 20\right)}{24}$

$= \frac{342 \cdot 420}{24}$

$= \frac{171 \cdot 420}{12}$

$= \frac{171 \cdot 210}{6}$

$= \frac{57 \cdot 210}{2}$

$= 57 \cdot 105$

$= 57 \cdot \left(100 + 5\right)$

$= 5700 + 57 \cdot \frac{10}{2}$

$= 5700 + 285$

$= \textcolor{b l u e}{5985}$