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Oct 11, 2017 at 15:40 history edited LRDPRDX CC BY-SA 3.0
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Mar 8, 2017 at 13:32 history edited LRDPRDX CC BY-SA 3.0
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Mar 7, 2017 at 9:38 history edited LRDPRDX CC BY-SA 3.0
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Mar 7, 2017 at 9:35 comment added LRDPRDX My solution in the context of $n$-dim cube is the following. Obviously, that there are $2^n$ vertices in $n$-dim cube. Then at each vertex choose $m$ $1$-dim edges which are part of $m$-dim cube. You will obtain $2^n C_{n}^{m}$. But you should divide this number by the number of how many times the same $m$ -dim cube was added in total sum which is $2^m$. Voilà, $Q_{n}^{m} = 2^{n-m}C_{n}^{m}.$
Mar 6, 2017 at 16:43 history edited LRDPRDX CC BY-SA 3.0
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Mar 6, 2017 at 16:35 vote accept LRDPRDX
Mar 6, 2017 at 10:12 history edited LRDPRDX CC BY-SA 3.0
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Mar 6, 2017 at 9:54 answer added T. Amdeberhan timeline score: 4
Mar 6, 2017 at 9:35 comment added Dima Pasechnik typically one would use a multivariate generating function for these kinds of problems.
Mar 6, 2017 at 8:30 history asked LRDPRDX CC BY-SA 3.0