However how do i prove 11 divides all of the possiblities? Use the fact that matrices commute under determinants 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.
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 You then take this entire sequence and repeat the process (abbabaab). 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.
$$\\overline{abba}$$ so it is equal to $$1001a+110b$$ and $1001$ and $110$ are 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. Are you required to make it wiht polar transformation
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