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Guichard-Combinatorics-Graph-Theory
· Problem 1.8.8
Fundamentals
Mathematics — Probability · stirling-numbers
Problem
Prove:
[
n
k
]
=
∑
i
=
k
−
1
n
−
1
(
n
−
i
−
1
)
!
(
n
−
1
i
)
[
i
k
−
1
]
\left[\begin{smallmatrix}n\\k\end{smallmatrix}\right] = \sum_{i=k-1}^{n-1} (n-i-1)! \binom{n-1}{i} \left[\begin{smallmatrix}i\\k-1\end{smallmatrix}\right]
[
n
k
]
=
∑
i
=
k
−
1
n
−
1
(
n
−
i
−
1
)!
(
i
n
−
1
)
[
i
k
−
1
]
.
Related theory
Stirling Numbers And Partition Functions
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