1. ## Combinations

2 more problems:

I insert the above in brackets into the combination formula nCk
which becomes: $\displaystyle \frac {n!}{(n-2)!(n-(n-2))!}=6$

solving for n I get 1.12 or 3.33, book gives the answer of 4.

I am incorrect for the second question(45) as well using the same method, my answer is 0/1/-7/-8/-9(approximated), book answer is 7.

2. ## Re: Combinations

Can you show how you get your answers? The first problem reduces to a quadratic equation. The second problem reduces to an equation of fourth degree, but after cancellation it also becomes a quadratic equation.

3. ## Re: Combinations

Originally Posted by emakarov
Can you show how you get your answers? The first problem reduces to a quadratic equation. The second problem reduces to an equation of fourth degree, but after cancellation it also becomes a quadratic equation.
I put both sides of the equations into a graphing calculator and use the intersections as "n".

&#40;n&#33;&#41;&#47;&#40;&#40;n-2&#41;&#33;&#40;n-&#40;n-2&#41;&#41;&#33;&#41;&#61;6 - Wolfram|Alpha

Looking back on my cellphones wolfram application history it was a transcription error after all.(for 43)

45:
&#40;&#40;n&#43;2&#41;&#33;&#47;&#40;4&#33;&#40;&# 40;n-2&#41;&#33;&#41;&#41;&#61;6&#40;&#40;n&#33;&#41;&# 41;&#47;&#40;2&#33;&#40;n-2&#41;&#33;&#41; - Wolfram|Alpha
apologies

4. ## Re: Combinations

Originally Posted by Greymalkin
2 more problems:

I insert the above in brackets into the combination formula nCk
which becomes: $\displaystyle \frac {n!}{(n-2)!(n-(n-2))!}=6$
$\displaystyle \binom{n+2}{4}=\frac{(n+2)(n+1)(n)(n-1)}{4!}$ and $\displaystyle 6\binom{n}{2}=6\frac{n(n-1)}{2!}~.$

Put those equal we get $\displaystyle n^2+3n-70=0~.$