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Thread: Factorising .

  1. #1
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    Factorising .

    Factorise each of the following below .

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

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

    (2) $\displaystyle \frac{10!}{7!3!}-\frac{8!}{5!4!}+\frac{7!}{3!2!}$
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by thereddevils View Post
    Factorise each of the following below .

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

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

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

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

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



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

    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: . $\displaystyle \frac{7!}{3!\,2!}\bigg[\frac{8}{4(5\!\cdot\!4\!\cdot\!3)} + 1\bigg] $

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




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

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


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



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