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$\binom{n}{k-1} + \binom{n}{k} = \binom{n+1}{k}$

So far I have gotten $\frac{n!}{(k-1)!\big(n-(k-1)\big)!} + \frac{n!}{ k! (n-k)!}$

But I quickly lose myself once I have to start making the denominator equal. Help would be appreciated.

Sorry for the duplication, I am new to the site and didn't quite know how to look up my question.

Goku
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    It's certainly possible to do this algebraically, but it's easier to think about what C(n,k-1) means - that is, what set it counts. – Milo Brandt Sep 30 '14 at 21:13
  • I have found this answer Here http://math.stackexchange.com/questions/86093/why-is-n1-choose-k-n-choose-k-n-choose-k-1 – Goku Sep 30 '14 at 21:16

2 Answers2

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$$\begin{array}{rcl}\binom{n}{k-1}+\binom{n}{k} & = & \frac{n!}{(n-k+1)!(k-1)!}+\frac{n!}{(n-k)!k!} \\ & \underbrace{=}_{(1)} & \frac{n!k}{(n-k+1)!k!}+\frac{n!(n-k+1)}{(n-k+1)!k!} \\ & = & \frac{n!(k+n-k+1)}{(n-k+1)!k!}\\ & = & \frac{n!(n+1)}{(n-k+1)!k!}\\ & = & \frac{(n+1)!}{(n-k+1)!k!}\\ & = & \binom{n+1}{k}\end{array}$$

In step $(1)$ we multiply by $k$ the numerator and denominator of the fraction $\frac{n!}{(n-k+1)!(k-1)!}$ to get $\frac{n!k}{(n-k+1)!k!}$ and we multiply by $n-k+1$ the numerator and denominator of the fraction $\frac{n!}{(n-k)!k!}$ to get $\frac{n!(n-k+1)}{(n-k+1)!k!}.$ Thus, both fractions have the same denominator and we can add them easily.

mfl
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  • Thank you very much for your help, Could you possibly next to each line show what you multiplied to the top and bottom to get that answer? – Goku Sep 30 '14 at 21:21
  • I have just done it. Please ask if you need some further clarification. – mfl Sep 30 '14 at 21:27
  • Ok as I was working it out in a longer way showing every step I get to the point where(forget the denominator for now sense it has been set) N * N! + N! is the numerator. How does this simply out to be N!(N+1) then to just (N+1)! – Goku Sep 30 '14 at 21:49
  • Just use $n!$ as common factor. That is, $(n+1)!=(n+1)\cdot n!=n\cdot n!+n\cdot 1=n\cdot n!+n.$ – mfl Sep 30 '14 at 21:55
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an other way: Let $A$ a set with $n+1$ element. Let $a\in A$.To count the subset with $k$ element of $A$, we can consider the subset that contain $a$ (and there is $\binom{n}{k-1}$ that contain $a$) and the subset that doesn't contain $a$, and there is $\binom{n}{k}$ subset.

Q.E.D.

idm
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