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June 16th, 2016, 11:17 PM  #1 
Newbie Joined: Jun 2016 From: Berlin Posts: 1 Thanks: 0  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 
June 19th, 2016, 03:29 AM  #2 
Newbie Joined: Jun 2016 From: Hong Kong Posts: 20 Thanks: 2 
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 nk+1 variables say n=4,k=2,nk+1=3 a<b<c=d,a<c=d<b,c=d<a<b,... $\displaystyle C_n^k (nk+1)!$ ways $\displaystyle n!+\sum_{k=2}^{n1}C_n^k (nk+1)!+1=1+n\sum_{k=2}^n \frac{n!}{k!}$ 

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numbers, permutation, relations 
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