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Guichard-Combinatorics-Graph-Theory
· Problem 1.8.9
Fundamentals
Mathematics — Probability · stirling-numbers
Problem
Use the previous exercise to prove
s
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n
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k
)
=
∑
i
=
k
−
1
n
−
1
(
−
1
)
n
−
i
−
1
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n
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i
−
1
)
!
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n
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1
i
)
s
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i
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k
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1
)
s(n,k) = \sum_{i=k-1}^{n-1} (-1)^{n-i-1}(n-i-1)! \binom{n-1}{i} s(i,k-1)
s
(
n
,
k
)
=
∑
i
=
k
−
1
n
−
1
(
−
1
)
n
−
i
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1
(
n
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i
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1
)!
(
i
n
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1
)
s
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i
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k
−
1
)
.
Related theory
Stirling Numbers And Partition Functions
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