# Thread: Factorising .

1. ## Factorising .

Factorise each of the following below .

(1) $\frac{8!}{4!5!}+\frac{7!}{3!2!}$

i got 2*7(31) but not sure whether it's correct .

(2) $\frac{10!}{7!3!}-\frac{8!}{5!4!}+\frac{7!}{3!2!}$

2. Hello,
Originally Posted by thereddevils
Factorise each of the following below .

(1) $\frac{8!}{4!5!}+\frac{7!}{3!2!}$

i got 2*7(31) but not sure whether it's correct .
Your answer is correct. But I'm not sure this is what you are asked to do, since you have to factorise these.

Note that $8!=8 \cdot 7!$, $4!=4 \cdot 3!$, $5!=5 \cdot 4 \cdot 3 \cdot 2!$

So $\frac{8!}{4!5!}=\frac{7!}{3!2!} \cdot \frac{8}{4 \cdot 5 \cdot 4 \cdot 3}$

Hence $\frac{8!}{4!5!}+\frac{7!}{3!2!}=\frac{7!}{3!2!} \left(\frac{2}{5 \cdot 4 \cdot 3}+1\right)=\frac{7!}{3!2!} \left(\frac{1}{30}+1\right)$

(2) $\frac{10!}{7!3!}-\frac{8!}{5!4!}+\frac{7!}{3!2!}$
Same here, 10!=10x9x8x7!, 7!=7x6x5x4x3!, 3!=3x2!, .........
Factor out $\frac{7!}{3!2!}$

3. Hello, thereddevils!

Factorise each of the following below.
Strange instructions . . .

If they truly want them factored, we would simply
. . factor out the greatest common factor ... and stop!

If they wanted them evaluated, they should have said so.

$(1)\;\;\frac{8!}{4!5!}+\frac{7!}{3!2!}$

i got 2*7(31) but not sure whether it's correct.
I'd like to see how you got that answer by factoring.

I would factor it like this: . $\frac{7!}{3!\,2!}\bigg[\frac{8}{4(5\!\cdot\!4\!\cdot\!3)} + 1\bigg]$

. . or maybe: . $\frac{7!}{4!\,5!}\bigg[8 + 4(5\!\cdot\!4\!\cdot\!3)\bigg]$

$(2)\;\;\frac{10!}{7!3!}-\frac{8!}{5!4!}+\frac{7!}{3!2!}$

$\frac{7!}{3!\,2!}\bigg[\frac{10\!\cdot\!9\!\cdot\!8}{(7\!\cdot\!6\!\cdot\ !5\!\cdot\!4)(3)} - \frac{8}{(5\!\cdot\!4)(4\!\cdot\!3)} + 1 \bigg]$

. . or: . $\frac{7!}{7!\,3!}\bigg[10\!\cdot\!9\!\cdot\!8 - 8(7\!\cdot\!6)(4) + (7\!\cdot\!6\!\cdot\!5\!\cdot\!4)(3)\bigg]$