I know if I have a set including $N$ members, the general formula for the number of distinct pairs of members is $$ N(N-1) $$ Now if I have two sets, $A$ and $B$, including $a$ and $b$ members respectively (with all of them different altogether), what is the general formula for the number of distinct pairs? for eaxmple $(a-1)(b-1)$?
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1You seem to be saying ‘non-distinct’ when you mean ‘distinct’ (or ‘non-equal’). That's the only way that $N(N-1)$ can be correct. I've submitted an edit with this in mind (and fixing some other grammar). – Toby Bartels Jan 02 '22 at 07:02
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1The reason for subtracting $1$ in the formula $N(N-1)$ is that the second time you pick, you can't pick the one that you picked first. But you start with $N$. So there's no reason to subtract $1$ at all in $(a-1)(b-1)$. If $A$ and $B$ are disjoint, and you want to pick an element of $A$ and an element of $B$, then there are $a$ ways to make the first choice and $b$ ways to make the second choice, so the total number of ways is simply $ab$. – Toby Bartels Jan 02 '22 at 07:24
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@TobyBartels Yes you're right, Thanks a lot. – Wisdom Jan 02 '22 at 07:48
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One often needs to clarify whether "pairs" means ordered pairs or unordered pairs. In your first case, to make the formula correct requires ordered pairs $(x,y)$ with $x,y$ belonging to your unnamed set of $N$ elements. In your second case, since $A,B$ are disjoint sets ("with all of them different altogether"), it would be the same count as any unordered pair containing one element of $A$ and one of $B$ could be uniquely ordered to make the element of $A$ come first. – hardmath Jan 03 '22 at 23:25