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 Apr 29 comment Eigenvector of matrix of equal numbers I still don't get how you got (-1, 1, 0) and (-1, 0, -1). I know it's the right answer, but I how do you get it from x1+x2+x3 = 0? Apr 29 comment Eigenvector of matrix of equal numbers I understand all of this. I still don't get how you'd get to (-1, 1, 0) and (-1, 0, -1) from $x_1 + x_2 + x_3 = 0$ Apr 28 comment Eigenvector of matrix of equal numbers So to make it in terms of $x_1$, you set $x_1$ to 1 and set once $x_2$ to zero, and in the other vector, $x_3$ to zero (the other would be -1 as the equation $x_1 = -x_2-x_3$ shows)? Apr 28 comment Eigenvector of matrix of equal numbers I know. This matrix has 3 eigenvalues, two of which are repeated, so I can generate the third from the second. But how would I find the second? (I already found the first corresponding to the single real-valued eigenvalue). This does not answer how to find it. Apr 10 comment Basketball Team Combinatorial @AustinMohr Why would the order matter though? 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 comment Basketball Team Combinatorial what does $2\times \binom{6}{3}\times \binom{4}{2}$ correspond to? 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: ${9 \choose 2} {7 \choose 1} + {9 \choose 1} {7 \choose 2} + {9 \choose 2} {5 \choose 1} + {9 \choose 1} {5 \choose 2} + {7 \choose 2} {5 \choose 1} + {7 \choose 1}{5 \choose 2}$ ? Sep 15 comment How to prove that $\sqrt 3$ is an irrational number? That's how you do it in Discrete Mathematics.