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bio website nayefcopty.com
location Amman, Jordan / Blacksburg, VA, USA
age
visits member for 2 years, 11 months
seen Apr 8 at 2:16

C, C++, iOS (Objective-C, Cocoa Touch), x86 Assembly, Java, Bash


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.