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I'm studying for my final for Statistics, and I want to understand literally every problem in my textbook (at least in the first 7 chapters).

One of the problems asks to show that ${n}\choose {k}$ $=$${n-1}\choose {k}$ $+$ ${n-1}\choose {k-1}$

I have the solution manual, but the steps it gives appears to skip a few important steps or leave out important clarifications.

Can someone please guide me step by step through this proof?

Siminore
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Grace C
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3 Answers3

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There are many ways to prove it. You may be looking for explanation for a specific way - in which case you need to post that way.

Here is perhaps a simple proof - the LHS i.e. $C(n, k)$ is the number of ways to select $k$ objects from $n$ distinct ones. Now let us say we have marked a specific object and want to count the selections with and without this object. You can select $k$ objects without the marked one in $C(n-1, k)$ ways, and including the marked one in $C(n-1, k-1)$ ways - thus both these sum up to the LHS.

Macavity
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  • I'm looking for lattice proof explanation. Actually, a video from combinatorics course in edx shows this proof but I was not satisfied with the explanation given there and I have some questions. Should I create a new thread for that? – mac07 Nov 23 '16 at 09:56
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    @mac07 a new thread with link to this one and showing your questions would be best. – Macavity Nov 23 '16 at 09:58
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I like think about it this way:

1) There are $n$ given objects, among which we have to choose $k$ elements. In the selection, 1 of the $n$ given elements never occur. This can be done in ${n-1 \choose k}$ ways.

2) Again, we have $n$ elements, in which 1 element always occurs. So, we are left to select $k-1$ elements from $n-1$ elements. This can be done in ${n-1 \choose k-1}$ ways, and we the number of ways we can choose the element that always occurs is 1.

Hence, the result.

Hawk
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Let's start with the right member of the equality : $${{n-1}\choose{k-1}}+{{n-1}\choose{k}} = \frac{(n-1)!}{((n-1)-(k-1))!(k-1)!}+\frac{(n-1)!}{((n-1)-k)!k!}$$ by definition

$$=\frac{(n-1)!}{(n-k)!(k-1)!}+\frac{(n-1)!}{(n-1-k)!k!}$$ simplifying $(n-1)-(k-1)=n-k$

$$=\frac{(n-1)!k}{(n-k)!k!}+\frac{(n-1)!(n-k)}{(n-k)!k!}$$ since $\frac{1}{(n-1-k)!}=\frac{n-k}{(n-k)(n-k-1)(n-k-2)...3*2*1}$

$$= \frac{(n-1)!k+(n-1)!(n-k)}{(n-k)!k!}$$

$$= \frac{(n-1)!(k+(n-k))}{(n-k)!k!}$$

$$= \frac{(n-1)!n}{(n-k)!k!}$$ $$= \frac{n!}{(n-k)!k!}$$ $$= {{n}\choose{k}}$$ by def

wibix
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