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Jul 16, 2011 at 6:22 comment added nvcleemp Uhu, you're completely right. blush
Jul 15, 2011 at 19:51 comment added Derrick Stolee Your counts of $v(v-1)$ pairs in $[v]$ and $6$ pairs in $[3]$ are really ordered pairs. The conditions are over unordered pairs, so you should say it is ${v \choose 2} / {3\choose 2}$. Since the factors of two cancel, you get the same number, but your reasoning isn't perfect.
Jul 15, 2011 at 15:50 vote accept Shay
Jul 15, 2011 at 15:06 comment added nvcleemp Please feel free to ask for clarifications where needed, because I'm terrible at using correct terminology.
Jul 15, 2011 at 13:09 history answered nvcleemp CC BY-SA 3.0