Skip to main content
MathOverflow

Return to Question

edited tags
Link
GH from MO
GH from MO
  • 105.4k
  • 8
  • 293
  • 398
improve title; added 2 characters in body
Source Link
Ryan Reich
Ryan Reich
  • 7.3k
  • 4
  • 37
  • 53

[combinatorics] show the formula Show that this ratio of factorials is always gives an integer

show the formula always gives an integer

$\frac{(2m)!(2n)!}{m!n!(m+n)!}$$$\frac{(2m)!(2n)!}{m!n!(m+n)!}$$

I don't remember where I read this problem, but it said this can be proved using a simple counting argument (like observing that $\frac{(3m)!}{m!m!m!}$ is just the number of ways of permuting m identical things of type 1, m of type-2 and m of type-3).

[combinatorics] show the formula always gives an integer

show the formula always gives an integer

$\frac{(2m)!(2n)!}{m!n!(m+n)!}$

I don't remember where I read this problem, but it said this can be proved using a simple counting argument (like observing that $\frac{(3m)!}{m!m!m!}$ is just the number of ways of permuting m identical things of type 1, m of type-2 and m of type-3).

Show that this ratio of factorials is always an integer

show the formula always gives an integer

$$\frac{(2m)!(2n)!}{m!n!(m+n)!}$$

I don't remember where I read this problem, but it said this can be proved using a simple counting argument (like observing that $\frac{(3m)!}{m!m!m!}$ is just the number of ways of permuting m identical things of type 1, m of type-2 and m of type-3).

Post Reopened by Douglas Zare, user6976, Joseph O'Rourke, Roland Bacher, Seva
Post Closed as "off topic" by Qiaochu Yuan, Steven Landsburg, Gerry Myerson, Fernando Muro, Andreas Blass
added 3 characters in body
Source Link
karan
karan
  • 383
  • 3
  • 6

show the formula always gives an integer

$\frac{(2m)!(2n)!}{m!n!(m+n)!}$

I don't remember where I read this problem, but it said this can be proved using a simple counting argument (like observing that $\frac{(3m)!}{m!m!m!}$ is just the number of ways of permuting m identical things of type 1, m of type-2 and m of type-3).

show the formula always gives an integer

$\frac{(2m)!(2n)!}{m!n!(m+n)!}$

I don't remember where I read this problem, but it said this can be proved using a simple counting argument (like observing that $\frac{(3m)!}{m!m!m!}$ is just the number of ways of permuting m identical things type 1, m of type-2 and m of type-3).

show the formula always gives an integer

$\frac{(2m)!(2n)!}{m!n!(m+n)!}$

I don't remember where I read this problem, but it said this can be proved using a simple counting argument (like observing that $\frac{(3m)!}{m!m!m!}$ is just the number of ways of permuting m identical things of type 1, m of type-2 and m of type-3).

Source Link
karan
karan
  • 383
  • 3
  • 6