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1
2010 PUMaC Individual Finals A1: combinatorial identity
2010 PUMaC Individual Finals A1: combinatorial identity
Source:
August 31, 2011
Problem Statement
Show that
∑
i
=
1
n
(
−
1
)
n
+
i
(
n
i
)
(
i
n
n
)
=
n
n
\displaystyle{\sum_{i=1}^{n}(-1)^{n+i}\binom{n}{i}\binom{in}{n} = n^{n}}
i
=
1
∑
n
(
−
1
)
n
+
i
(
i
n
)
(
n
in
)
=
n
n
.
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