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Show that:

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

I think I am supposed to use the number of Combinations formula:

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

I tried to use the formula on the LHS and got this:

$\frac{(n+1)!}{2(n-1)}$

2 Answers2

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You can also argue "combinatorically":

  • LHS: all selections of $2$ items from $n+1$ items: $\binom{n+1}{2}$
  • RHS: fix the first $n$ items of the $1 \ldots n \; (n+1)$ items, take all selections of $2$ out of these $n$ items and all selections consisting of item $(n+1)$ and one of the first $1 \ldots n$ items: $\binom{n}{2} + n$
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I got this solution:

$ \binom{n+1}{2} = \frac{(n+1)!}{2(n-1)!} = \frac{(n+1)n!}{(n-1)2!(n-2)!} = \frac{n!}{2(n-2)!} * \frac{n+1}{n-1} = $

$ = \binom{n}{2} * (1 + \frac{2}{n-1}) = \binom{n}{2} + \frac{2n!}{(n-1)2(n-2)!} = \binom{n}{2} + n $

However i don't understand this part:

$ \binom{n}{2} * \frac{n+1}{n-1} = \binom{n}{2} * (1 + \frac{2}{n-1}) $

How can this be true?