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1970 Swedish Mathematical Competition
6
(n - m)!/m! <= (n/2 + 1/2)^{n-2m} if 0< 2m < n
(n - m)!/m! <= (n/2 + 1/2)^{n-2m} if 0< 2m < n
Source: 1970 Swedish Mathematical Competition p6
March 21, 2021
inequalities
algebra
Problem Statement
Show that
(
n
−
m
)
!
m
!
≤
(
n
2
+
1
2
)
n
−
2
m
\frac{(n - m)!}{m!} \le \left(\frac{n}{2} + \frac{1}{2}\right)^{n-2m}
m
!
(
n
−
m
)!
≤
(
2
n
+
2
1
)
n
−
2
m
for positive integers
m
,
n
m, n
m
,
n
with
2
m
≤
n
2m \le n
2
m
≤
n
.
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