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Simplify $\binom{m+n}{2} - \binom{m}{2} - \binom{n}{2}$.


I'm confused on how to approach this problem. I can't think of any counting argument that will help me, and any 1-1 correspondence. All solutions are appreciated.

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2 Answers 2

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HINT: You have $m$ men and $n$ women in a room. $\binom{m+n}2$ is the number of ways to pick two people in the room. $\binom{m}2$ is the number of ways to pick two of the men, and $\binom{n}2$ is the number of ways to pick two of the women. If you remove those possibilities, what’s left?

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Algebraic approach:
Use ${k \choose 2} = \frac{k(k-1)}{2}$ and simplify.

Combinatorial approach:
Suppose you have $m$ distinct red balls and $n$ distinct green balls. You want to choose $2$ of the $m+n$ total balls but you don't want them to both be red and you don't want them to both be green.

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  • $\begingroup$ Is $mn-m-n$ the correct answer? $\endgroup$
    – Yuna Kun
    Commented Oct 7, 2016 at 19:50
  • $\begingroup$ How did you get that? Always show your work. $\endgroup$ Commented Oct 7, 2016 at 19:51
  • $\begingroup$ I used the algebra way and got $\frac{m^2-m-n+n^2+2mn-m^2-m-n^2-n}{2}.$ $\endgroup$
    – Yuna Kun
    Commented Oct 7, 2016 at 19:52
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    $\begingroup$ I just realized my error! I had to change the signs! So is $mn$ the correct answer? $\endgroup$
    – Yuna Kun
    Commented Oct 7, 2016 at 19:53
  • $\begingroup$ If you want two balls and they cannot be the same color, you need $1$ of the $m$ red balls and $1$ of the $n$ green balls. Yes, there are $m \cdot n$ ways to make that selection. $\endgroup$ Commented Oct 7, 2016 at 20:01

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