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I'd like to count the number of pairs of subsets $A,B$ of $\{1, 2, \dots, n\}$ such that $|A|=|B|=k$ and if $A=\{a_1, a_2, \dots, a_k\}$ with $a_1 < a_2 < \cdots < a_k $ and $B=\{b_1, b_2, \dots, b_k\}$ with $b_1 < b_2 < \cdots < b_k $, then $ a_i = b_i $ for $i=1, 2, \dots, m$. Ideally, I'd like to find a closed form expression (without sums) involving $ n $, $ k $, and $ m $.

Does anyone know any references for tackling these sort of counting problems? Thanks!

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The answer with a summation is $$\sum_{i=m}^{n-k+m} \binom{i-1}{m-1}\binom{n-i}{k-m}^2.$$ The interpretation is $i=a_m=b_m$. Choose $\{a_1,\dots,a_{m-1}\}$ in $\binom{i-1}{m-1}$ ways, then choose $\{a_{m+1},\dots,a_k\}$ and $\{b_{m+1},\dots,b_k\}$ in $\binom{n-i}{k-m}^2$ ways.

I think it is very likely that no closed form exists. I applied the methods of the book A = B to this summation, first using Zeilberger's algorithm to find a linear recurrence satisfied by your summation, then Petkovšek's algorithm to determine if this recurrence had a closed form, and found none. This proves there are no hypergeometric closed forms to your summation, which means there is no closed form involving arithmetic operations, exponentiation with constant bases, factorials, and binomial coefficients.

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  • $\begingroup$ This is perfect! Thank you so much. $\endgroup$
    – mr_snazzly
    May 15 at 17:00

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