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EDITED Let $d > 0$ be a fixed integer. Let $m_i = \binom{d}{i}$ so that $\sum_{i=0}^d m_i = m = 2^{d}$

How fast does $S_d = \prod_{i=0}^d m_i!$ grow in relation to $m!$ ? is $S_d = O(m!)?$ How about $o(m!)?$ or is it actually lower bounded by $m!$?

Also, more generally, the answer to the question stated by Zev below, and the equivalent when 'max' is replaced by 'min' would also be of interest.

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It's not clear what you mean by this question. Do you mean to ask, how fast does $$\max_{\substack{m_1,\ldots,m_n\geq 0\\ \sum_i m_i=m}}\left\{\prod_i m_i!\right\}$$ grow as a function of $m$? –  Zev Chonoles Mar 1 '13 at 5:12
    
Are your $m_i$ partitions or compositions? If $m=6$, do we count $4,2$ separately from $2,4$? It makes quite a difference. –  Ross Millikan Mar 1 '13 at 5:58
    
Sorry, editing the question now for clarity. –  rhl Mar 1 '13 at 6:32
    
$S_d$ is $O(m!)$; in fact it's less than $m!$: It has as many factors as $m!$ (namely $m$), but some of them are smaller. That also answers your question whether it's lower bounded by $m!$ in the negative. The only non-trivial question here is whether it's $o(m!)$; you might want to focus the post on that. –  joriki Mar 1 '13 at 7:48
    
Well, just because it is O(m!) doesn't mean that it is not $\Omega(m!)$ as well.. I suspect that it still is, because some $m_i \geq m/d$ so you must have at least $(m/d)!$ –  rhl Mar 1 '13 at 9:06

1 Answer 1

up vote 1 down vote accepted

Decreasing all but the largest $m_i$ to $1$ while correspondingly increasing the largest $m_i$ replaces smaller factors by larger factors, and thus increases the product. The result is $(m-d)!$, which is $o(m!)$, so $S_d$ is $o(m!)$.

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