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The question is in the title. I have no idea how to do this one. I was trying to rewrite $n^2$ as $\binom {n}{1} \times \binom {n}{1}$. I was going to say the story is, suppose there are $n$ boys and $n$ girls. How many ways there are to pick $1$ boy and $1$ girl, which is the LHS. But then I don't have any idea how to prove the RHS.

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    $\begingroup$ I see it geometrical: are of square equals sum of areas of two rectangles $n\cdot (n-1)$ and $n\cdot 1$. $\endgroup$
    – zkutch
    Oct 18 at 23:50
  • $\begingroup$ I note $n(n-1) = 2 \binom{n}{2}$. $\endgroup$ Oct 18 at 23:53
  • $\begingroup$ @zkutch omg that's so clever! $\endgroup$ Oct 19 at 0:26
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One idea: consider counting ordered pairs of integers $(a, b)$ in the range $\{1, \ldots, n\}$. Count the cases $a = b$ and $a \neq b$ separately.

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If we have two people that need to choose from $n$ items (not necessarily distinct),

then there are $n \cdot n = n^{2}$ possibilities.

If we instead count the distinct possibilities first we subtract one for the second person's choice, which gives $n(n-1)$.

Note that since we have $n$ items, there are exactly $n$ possibilities where they do choose the same item.

These are independent choices with the union being the first counting method:

i.e. Distinct + Not distinct = Not necessarily distinct.

Hence we add them and get $n^{2}=n(n-1)+n$.

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  • $\begingroup$ thank you! I has this thought at first but I denied it for some reason. $\endgroup$ Oct 19 at 0:24

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