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How many ways can we get 2 as and 2 bs from aabb? If I do this manually, it's clear to me the answer is 6, aabb abab abba baba bbaa baab Which is the same as \mathchoice((((4 2\mathchoice)))) \mathchoice ((((4 2 \mathchoice)))) But I don't really understand why this is true? How is this supposed to be done without brute forcing the question?
How to calculate total combinations for AABB and ABBB sets? Although both belong to a much broad combination of N=2 and n=4 (AAAA, ABBA, BBBB ), where order matters and repetition is allowed, both can be rearranged in different ways: First one: AABB, BBAA,
$A^2=AB+BA$. Prove that $\\det(AB-BA)=0$ Let A, B A, B be two 3 × 3 3 × 3 matrices with complex entries, such that A2 = AB + BA A 2 = A B + B A Prove that det(AB − BA) = 0 det (A B B A) = 0 Nice problem, and I want to find a solution AB − BA = A2 − 2BA = (A − 2B)A A B B A = A 2 2 B A = (A 2 B) A so if |A| = 0 | A | = 0 we have done, if |A| ≠ 0 | A | ≠ 0 I can't prove
random variables - probability of playing music player on shuffle and . . . 1 How many sequences of 4 songs are there where no song plays twice in a row? If we label the songs {A, B, C, D}, then examples are ABCD and ABAB but not ABBA For this problem I just thought the answer was (4^4) = 256 Does this make sense? 2
The commutator of two matrices - Mathematics Stack Exchange The commutator [X, Y] of two matrices is defined by the equation $$\begin {align} [X, Y] = XY − YX \end {align}$$ Two anti-commuting matrices A and B satisfy $$\begin {align} A^2=I \qu
How to show that - Mathematics Stack Exchange Let A A and B B be two 3 × 3 3 × 3 matrices with complex entries such that A2 = AB + BA A 2 = A B + B A Prove that det(AB − BA) = 0 det (A B B A) = 0 (Is the above result true for matrices with real entries?)