However how do i prove 11 divides all of the possiblities? 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 In digits the number is $abba$ with $2 (a+b)$ divisible by $3$. Given two square matrices $a,b$ with same dimension, what conditions will lead to this result Or what result will this condition lead to I thought this is a quite.
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. Use the fact that matrices commute under determinants There must be something missing since taking $b$ to be the zero matrix will work for any $a$.
This is nice work and an interesting enrichment I realized when i had solved most of it that the op seems to know how to compute the generating function but is looking for a way to extract the coefficients using pen and paper. For example a palindrome of length $4$ is always divisible by $11$ because palindromes of length $4$ are in the form of $$\\overline{abba}$$ so it is equal to $$1001a+110b$$ and $1001$ and $110$ are
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