darksky
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 Apr 10 comment Basketball Team Combinatorial So the problem with my above comment over counts. True? Apr 10 comment Basketball Team Combinatorial Ah! got it. Thank you! Apr 10 comment Basketball Team Combinatorial I don't understand why my answer includes the possibility of not using the two free positioned players thrice. It only does it once, in the last adding expression. The first three expressions assume that 2 are forwards, then 1 is forward and 1 is guard, and then 2 are guards. Apr 10 comment Basketball Team Combinatorial I believe the question does not require those two players to play. Your final answer assumes both DO play. Correct? Apr 10 comment Basketball Team Combinatorial I like this approach. Would: $\binom{8}{3}\binom{4}{2} + \binom{7}{3}\binom{5}{2} + \binom{6}{3}\binom{6}{2} + \binom{6}{3}\binom{4}{2}$ be equivalent? This assumes both are forwards, then 1 is forward and 1 is guard, then 2 are guards, then none. If not, would would this be wrong? Apr 10 comment Basketball Team Combinatorial Would: $\binom{8}{3}\binom{4}{2} + \binom{7}{3}\binom{5}{2} + \binom{6}{3}\binom{6}{2} + \binom{6}{3}\binom{4}{2}$ be equivalent? This would include not using him at all once. Apr 10 awarded Student Apr 10 comment Basketball Team Combinatorial what does $2\times \binom{6}{3}\times \binom{4}{2}$ correspond to? Apr 10 asked Basketball Team Combinatorial Apr 4 comment Picking 3 Books Combinatorics Question Would this count it properly: ${9 \choose 2} {7 \choose 1} + {9 \choose 1} {7 \choose 2} + {9 \choose 2} {5 \choose 1} + {9 \choose 1} {5 \choose 2} + {7 \choose 2} {5 \choose 1} + {7 \choose 1}{5 \choose 2}$ ? Apr 4 comment Picking 3 Books Combinatorics Question So: ${9 \choose 2} {7 \choose 1} + {9 \choose 1} {7 \choose 2} + {9 \choose 2} {5 \choose 1} + {9 \choose 1} {5 \choose 2} + {7 \choose 2} {5 \choose 1} + {7 \choose 1}{5 \choose 2}$ ? Apr 4 awarded Scholar Apr 4 asked Picking 3 Books Combinatorics Question Sep 15 awarded Supporter Sep 15 comment How to prove that $\sqrt 3$ is an irrational number? That's how you do it in Discrete Mathematics.