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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,
How many $4$-digit palindromes are divisible by $3$? How many 4 4 -digit palindromes are divisible by 3 3? I'm trying to figure this one out I know that if a number is divisible by 3 3, then the sum of its digits is divisible by 3 3 All I have done is listed out lots of numbers that work I haven't developed a nice technique for this yet
How many words of length - Mathematics Stack Exchange {AABB, ABAB, ABBA, BBAA, BAAB, BABA} {A A B B, A B A B, A B B A, B B A A, B A A B, B A B A} So, I need to know what this w(n) = w(ni,nj) w (n) = w (n i, n j) is Some references on aproaches on how to solve this kind of problems and related algorithms to generate such words would be apreciated Thanks
(Jacobson Basic Algebra I) Exercise 1. 2. 7 (monoid inverse exercise) I hadn't made the leap to realize that abba = 1 means that a is left right invertible and therefore has an inverse, it's rather clever Also thank you for the minimal counter example, it's very accessible for an abstract algebra learner like me!
elementary number theory - Divisibility Tests for Palindromes . . . The 4 4 -digit palindrome abba a b b a is divisible by 101 iff a = b a = b The 5 5 -digit palindrome abcba a b c b a is divisible by 101 iff c = 2a c = 2 a The 6 6 -digit palindrome abccba a b c c b a is divisible by 101 iff a + b = c a + b = c The 7 7 -digit palindrome abcdcba a b c d c b a is divisible by 101 iff d = 2b d = 2 b