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How to get from $\frac{n(n+1)}{2}$ to $n + \binom{n}{2}$?

I tried some transforming, but my main problem is that I don't know how to get to $n!$ and $(n-2)!$ from the original term.

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  • $\begingroup$ Doesn't putting $^nC_2$ as $\frac{n(n-1)}{2}$ serve your purpose? $\endgroup$ Jun 23, 2014 at 7:00

5 Answers 5

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We know that $$\binom{n}{r}+\binom{n}{r-1}=\binom{n+1}{r}.$$ By this we have $$n+\binom{n}{2}=\binom{n}{2}+\binom{n}{1}=\binom{n+1}{2}$$

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Just for kicks and giggles (probably not of help to the OP), here is a combinatorial proof.

Let $A$ be a set and $*\not\in A$. Let $\binom{A}{2}$ denote the set of $2$-subsets of $A$. Define a map

$$A\times(A\sqcup\{*\})\to A\sqcup\binom{A}{2}$$

given by $(x,y)\mapsto\{x,y\}$ when $x\ne y$, and $(x,x)\mapsto x$ and $(x,*)\mapsto x$. One may verify that this mapping is $2$-to-$1$, hence taking cardinalities yields $n(n+1)/2=n+\binom{n}{2}$, where $n=|A|$.

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Hint $$\binom n2 = \frac{n!}{(n-2)!2!} = \frac{n(n-1)}2 $$

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Here's a combinatorial proof:

Suppose we have n boys and 1 girl in a room and we have to select 2 people out of them.

Then we can do that in $\binom{n+1}{2}=\frac{n(n+1)}{2}$ ways since we have n+1 people to choose from.

Counting in another way. Case 1: There is a girl in our selection. Then we can do that n ways. Case 2: There is no girl in our selection. We can do that in $\binom{n}{2}$ ways since we have n boys to choose from.

Since these two must be equal. We have: $$\frac{n(n+1)}{2}=n+\binom{n}{2}$$

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If you have a more than casual encounter with the expression $\frac{n(n+1)}2$, then it is useful to know that it is equal to $\binom{n+1}2$, and often better written as such. Here this leads to observing that your equation is just a particular instance of Pascal's recursion (the additive recursion that let's you construct Pascal's triangle without any multiplications): $$ \frac{n(n+1)}2 = \binom{n+1}2 = \binom n1+\binom n2 = n +\binom n2 $$ since of course $\binom n1=n$ for all$~n$.

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