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June 17th, 2016, 12:17 AM   #1
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permutation of numbers and relations

How can I find the permutation for n different numbers and two relations (equal and less than)?
for example for n=2 (2 different numbers), we have 3 permutation:
a = b
a < b
b < a
and for n=3 (3 different numbers), we have 13 permutation:
a = b = c
a = b < c
a < b = c
a < b < c
a = c < b
a < c < b
b = c < a
b < c = a
b < c < a
b < a < c
c < b = a
c < b < a
c < a < b
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June 19th, 2016, 04:29 AM   #2
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when all variables equal
1 way

when all variables differ
a<b<c,a<c<b,b<a<c,...
n! ways

when k variables equal
first,choose k variables from n variables
second,permute n-k+1 variables
say n=4,k=2,n-k+1=3
a<b<c=d,a<c=d<b,c=d<a<b,...
$\displaystyle C_n^k (n-k+1)!$ ways

$\displaystyle n!+\sum_{k=2}^{n-1}C_n^k (n-k+1)!+1=1+n\sum_{k=2}^n \frac{n!}{k!}$
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