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Prove:

$\binom{n}{r-1} + 2\binom{n}{r} + \binom{n}{r+1} = \binom{n+2}{r+1}$

This was one of the questions that my professor gave me as extra practise however I want to know a more efficient way of getting to the answer as my method was very long.

Thanks in advance and good luck!

Oke so I understand most of this now however, I am still not entirelly sure why:

$\bigg[\binom{n}{r-1}+\binom{n}{r}\bigg] = \binom{n+1}{r}$

Can anyone explain this to me?

Pauly
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3 Answers3

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HINT:

$n \choose {r-1}$ $+$ $n \choose r$ $=$ ${n+1} \choose r$

Shobhit
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say that you have $n+2$ people and you want to choose $r+1$ out of them. Let's say that among these $n+2$ persons are 2 that you're interested in - A and B. now the RHS tells you all the ways of choosing r+1 persons out of a group of n+2. The LHS expresses the same number but with respect to what happens to A and B. Either you pick both A and B and you just need to pick the remaining $r-1$ out of $n$, OR exactly one of them is chosen - so you choose which one (in $2$ ways) and then you pick the remaining r persons out of n OR none of them is chosen so you pick r+1 out of n.

mm-aops
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Using your previous qustion Prove that $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1} $

$$\binom{n}{r-1} + 2\binom{n}{r} + \binom{n}{r+1} = \bigg[\binom{n}{r-1}+\binom{n}{r}\bigg] + \bigg[\binom{n}{r}+\binom{n}{r+1}\bigg]=\\\binom{n+1}{r}+\binom{n+1}{r+1}=\binom{n+2}{r+1}$$

Ashot
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