MathDB
Problems
Contests
National and Regional Contests
PEN Problems
PEN A Problems
26
26
Part of
PEN A Problems
Problems
(1)
A 26
Source:
5/25/2007
Let
m
m
m
and
n
n
n
be arbitrary non-negative integers. Prove that
(
2
m
)
!
(
2
n
)
!
m
!
n
!
(
m
+
n
)
!
\frac{(2m)!(2n)!}{m! n!(m+n)!}
m
!
n
!
(
m
+
n
)!
(
2
m
)!
(
2
n
)!
is an integer.
floor function
inequalities
induction
ratio
Divisibility Theory