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Prove that ${n \choose k} + {n \choose k+1} = {n+1 \choose k+1}$. I can prove it using numeric examples, but i need a broader proof. I think it might have to do with Pascal's Triangle.

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    $\begingroup$ Just write the binomial coefficient as factorials and rearrange terms to show the equality, i.e. $\binom{n}{k}=\frac{n!}{k!(n-k)!}$. $\endgroup$
    – Masacroso
    Oct 29, 2016 at 1:52

1 Answer 1

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Argumentative Method:

We know that the number of ways to choose $k$ objects from $n$ objects is $\binom{n}{k}$. Let us add 1 more object into the group. Now we have to choose $k+1$ objects from $n+1$ objects.

Lets say we include the newest item in the group. Then the number of ways to choose from the remaining $n$ objects is $\binom{n+1-1}{k+1-1} = \binom{n}{k}$.

Lets say we do not include the newest item in the group. Then the number of ways to choose from the remaining $n$ objects is $\binom{n+1-1}{k+1} = \binom{n}{k+1}$.

Since there are no overlaps between the 2 cases, $\binom{n+1}{k+1} = \binom{n}{k} + \binom{n}{k+1}$.

Mathematical Method:

$$ \begin{align*} \binom{n}{k} + \binom{n}{k+1} & = \frac{n!}{k!(n-k)!} + \frac{n!}{(k+1)!(n-k-1)!} \\ & = \frac{n!(k+1)}{(k+1)!(n-k)!} + \frac{n!(n-k)}{(k+1)!(n-k)!}\\ & = \frac{n!(n-k+k+1)}{(k+1)!(n-k)!}\\ & = \frac{n!(n+1)}{(k+1)!(n-k)!}\\ & = \frac{(n+1)!}{(k+1)!(n-k)!}\\ & = \binom{n+1}{k+1} \end{align*} $$

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