Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
@BrianM.Scott Yes, of course. –  Sh.N. May 22 '12 at 9:15
add comment

1 Answer

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.