How many ways can we get 2 as and 2 bs from aabb? Because abab is the same as aabb I was how to solve these problems with the blank slot method, i e _ _ _ _ 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 $$\binom {4} {2}$$ But I don't really understand why this is true? How is this supposed to be done without brute forcing the
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,
elementary number theory - Divisibility Tests for Palindromes . . . I've found and proven the following extensions to palindromes of the usual divisibility rules for 3 and 9: A palindrome is divisible by 27 if and only if its digit sum is A palindrome is divisible by 81 if and only if its digit sum is This doesn't straightforwardly extend to 243 As an example, using this you can immediately see the smallest palindromic multiple of 81 is 999999999, and the
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