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Prove combinatorially that

$$ \dbinom{p+q}{2} - \dbinom{p}{2} - \dbinom{q}{2} = \dbinom{p}{1} \times \dbinom{q}{1} $$

Can we also generalize this?

I thought of choosing $2$ objects out of $p+q$ objects and subtracting the cases where we choose from $p$ objects and from $q$ objects, but it didn't help.

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You have two boxes, while one has $p$ balls and another has $q$ balls, then: $$\binom {p+q}2=\text{the way you choose 2 balls from the two boxes}$$ $$\binom p2=\text{the way you choose the two from the $p$ box}$$ $$\binom q2=\text{the way you choose the two from the $q$ box}$$ Thus, $$\binom {p+q}2-\binom p2-\binom q2$$ $$=\text{the way you choose the two balls from the two boxes}-\text{the case that they are both from the $p$ box}-\text{the case that they are both from the $q$ box}$$ $$=\text{the way you choose the two balls from the two boxes, with not both of them are from the same box}$$ $$=\text{the way you choose one ball from each box respectively}$$$$=\binom p1\times\binom q1$$

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