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How to prove that $d$-dimensional hypercube has $2^{d-m}C_{m}^{d}$ $m$-dimensional hypercubes on his boundary?

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@BrianM.Scott Yes, of course. – Sh.N. May 22 '12 at 9:15
up vote 1 down vote accepted

Let $$B^d=\big\{\langle b_1,\dots,b_d\rangle:b_k\in\{0,1\}\text{ for }k=1,\dots,d\big\}$$ be the vertex set of the $d$-dimensional hypercube $C_d$. Suppose that $0\le m\le d$. Choose any $F\subseteq[d]$ of cardinality $d-m$, and for each $k\in F$ fix $a_k\in\{0,1\}$; then $$\big\{\langle b_1,\dots,b_d\rangle\in B^d:b_k=a_k\text{ for each }k\in F\big\}$$ is the vertex set of an $m$-dimensional ‘face’ of $C_d$, and every $m$-dimensional face of $C_d$ arises in this way. There are $\binom{d}{d-m}=\binom{d}m$ ways to choose the set $F$, and there are $2^{d-m}$ ways to choose the values $a_k$ for $k\in F$, so $C_d$ has $2^{d-m}\binom{d}m$ $m$-dimensional faces.

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