# Permutations, from Duncan Coker

March 13, 2015 |

I am reading a math book now and came across a section regarding permutations. I thought it relevant in looking at various markets and the possible combinations like a ranking. If you had 10 markets for example, I was curious how many possibilities there are of those 10 when the order is noted. Turns out there are 3.6 million combinations for just those 10 markets. The formula is P(n,r) = n! / (n - r)! where N is a set of items and R a sub set of selected items. In my example both N and R are 10. Intuitively I would have guessed much lower and shows how my brain at least is not very good estimated very large numbers. Math experts please educate me if I have erred.

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1. Yan Kan on March 13, 2015 5:05 pm

Assuming n items, for n > 1 belonging to N
and that each of them can appear only once in the combination,
I would say that there are (n^2)-n possible combinations.

2 items: 2 {1,2 ; 2,1}
3 items: 6 {123 ; 132 ; 213 ; 232 ; 312 ; 321}
4 items:12 {1234 ; 1243 ; 1324 ; 1342 ; 1423 ; 1432 ; 2134 ; 2143 ; 2314 ; 2341 ; 2413 2431}
10 items: 90 possible combination

2. Yan Kan on March 13, 2015 5:18 pm

I found 43 769 7000 possible combinations.
Using nPr=n!

3. Yan Kan on March 13, 2015 5:20 pm

My first comment’s bullshit sorry.