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Use the defintion of binomial theorem to prove the identy. $$\binom{n}{k} = \binom{n-1}{k}+\binom{n-1}{k-1}$$

The definition of the binomial theorem $$\binom{n}{k} = \frac{\prod_{i=0}^{k-1}(n-i)}{k!}$$

$$\binom{n-1}{k} = \frac{\prod_{i=0}^{k-1}(n-1-i)}{k!}$$

$$\binom{n-1}{k-1} = \frac{\prod_{i=0}^{k-2}(n-1-i)}{k-1!}$$

$$\frac{\prod_{i=0}^{k-1}(n-1-i)}{k!}+\frac{\prod_{i=0}^{k-2}(n-1-i)}{k-1!}$$

Then I come to the result

$$\frac{\prod_{i=0}^{k-2}(n-1-i)(n)}{k-2!(k-1)k}$$

$$\frac{\prod_{i=0}^{k-2}(n-1-i)}{k!} \cdot n$$

What can I do to make this result into

$$\frac{\prod_{i=0}^{k-1}(n-i)}{k!}$$

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    Your definition makes things increidibly cumbersome. Why not take the much easier to manage $$\binom nk:=\frac{n!}{k!(n-k)!}\;?$$ Of course, both definitions are exactly the same... – 2017-01-22
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    @ DonAntonio I know, but this is part of the requirement of my assignment. – 2017-01-22
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    @t Fair enough. Then, **first**, show your definition is identical as mine... :) – 2017-01-22
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    @ I can handle the formula you gave easily as it is just simple cancellation. It's a proof base question, so I have to use the definition. – 2017-01-22
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    As written above: show first your definition and mine are just the same, and then use the other one... – 2017-01-22

4 Answers 4

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Open directly the right hand side:

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

$$=\frac{(n-1)!}{(k-1)!(n-k-1)!}\left[\frac1k+\frac1{n-k}\right]=\frac{{(n-1)!}}{\color{red}{(k-1)!}\color{green}{(n-k-1)!}}\cdot\frac n{\color{red}k\color{green}{(n-k)}}=$$$${}$$

$$=\frac{n!}{k!(n-k)!}=\binom nk$$

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I think the DonAntonio's idea is the best way to approach the problem, but if you really want finish your idea then take a look. You missed the index at the second term (in red).

$$\binom{n-1}{k} = \frac{\prod_{i=0}^{k-1}(n-1-i)}{k!}$$

$$\binom{n-1}{k-1} = \frac{\prod_{i=0}^{\color{red}{k-2}}(n-1-i)}{(k-1)!}$$

So,

$$\frac{\prod_{i=0}^{k-1}(n-1-i)}{k!}+\frac{\prod_{i=0}^{k-2}(n-1-i)}{(k-1)!}=\frac{\prod_{i=0}^{k-1}(n-1-i)}{k!}+\frac{k\cdot\prod_{i=0}^{k-2}(n-1-i)}{k!}=\\ =(n-1-(k-1)+k)\cdot \frac{\prod_{i=0}^{k-2}(n-1-i)}{k!}=n\cdot\frac{\prod_{i=0}^{k-2}(n-1-i)}{k!}=\frac{\prod_{i=0}^{k-1}(n-i)}{k!}={n \choose k}$$

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    Thanks, I agree that Don's method is the best, however, I just want to practice the product Pi operation. I got the same result as you and the mistakes in my write up earlier was typo. However, would you mind how did you incorporate the n into the final term? that was the part I didn't get it. – 2017-01-22
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    Just see that $n\prod_{i=0}^{k-2}(n-1-k)=n(n-1)(n-2)...(n-k+1)=\prod_{i=0}^{k-1}(n-k)$. Is it clear? – 2017-01-22
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    @ Arnaldo Perfect! – 2017-01-22
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$\binom {n}{k}$ is the co-efficient of $x^k$ in $(1+x)^n$, which is equal to the co-efficient of $x^k$ in $$(1+x)^{n-1}+(1+x)^{n-1}x$$ which is the sum of the co-efficient of $x^k$ in $(1+x)^{n-1}$ and the co-efficient of $x^{k-1}$ in $(1+x)^{n-1},$ which is $\binom {n-1}{k}+\binom {n-1}{k-1}.$

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it cleat that $$\frac{\prod_{i=0}^{k-1}(n-1-i)}{k!}+\frac{\prod_{i=0}^{k-2}(n-1-i)}{k-1!}=\frac{n\prod_{i=0}^{k-2}(n-1-i)}{k!}$$ put (j=i+1) we obtain $$n\frac{\prod_{j=1}^{k-1}(n-j)}{k!}=\frac{\prod_{j=0}^{k-1}(n-j)}{k!}$$ which the result.