3
$\begingroup$

This question already has an answer here:

Show that $\binom{2m}{m} = \Theta\left(\frac{2^{2m}}{\sqrt{m}}\right)$ without using Stirling's approximation.

$\endgroup$

marked as duplicate by Dilip Sarwate, joriki, Thomas Andrews, Emily, Stefan Hansen Mar 13 '13 at 15:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ It would be neat if we could find a combinatorial proof that $m\binom{2m}{m}^2 \leq 2^{4m}$. Clearly, $\binom{2m}{m}^2$ counts a particular class of subsets of size $2m$ in the set of size $4m$. $\endgroup$ – Thomas Andrews Mar 13 '13 at 16:00
1
$\begingroup$

Not a full answer, but some suggestions: the numerator can be rewritten as $$ (2m)!=1 \cdot 2 \cdot \ldots \cdot 2m=1\cdot 3\cdot \ldots \cdot (2m-1)(2 \cdot 4 \cdot \ldots \cdot 2m=(2-1) \cdot (4-1)\cdot \ldots \cdot (2m-1) (2 \cdot 1) \cdot (2 \cdot 2) \cdot (2 \cdot 3) \cdot \ldots \cdot 2m \leq 2^{m} m! \cdot 2^m m!=2^{2m} (m!)^2 $$ Then you remain with $2^{2m}$ as an upper bound. Can you handle from here?

$\endgroup$

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