what is the probability that a randomly selected permutation of 1,2,3,...n?

Question

has the "2" somewhere between the "1" and the "3"?

guess what wannabees, I know, you tell. this is not a stupid homework problem

actually, this WILL be a homework problem if and only if someone gets it right.

i want the correct answer gjmb1960
yours is incorrect.

wrong answerer 1 flavorxo
wrong answerer 2 Christicide
already wrong answerer 3

Answer 1

Hi Scott....I believe the answer = 1/3

For the probability denominator, we simply have: n!

We need to determine the numerator:

First take the number of ways of choosing 3 out of n = nC3
These chosen 3 will be our numbers 1,2 and 3.
For each of the nC3 ways, we have 6 permutations of 1,2,3 but only one 3rd of these, i.e. two permutations, will have 2 in the middle. Also, for each of these nC3 ways, for the remaining n-3 positions there are (n-3)! permutations.
So the numerator will be 2 x nC3 x (n-3)!; and so the desired probability will be:

2 x nC3 x (n-3)! / n! = 2 x n(n-1)(n-2)(n-3)!/ (3x2xn!) = 1/3

Answer 2

total number of permutations is N !

total number of succesfull permutations is (N-2) !
(treat the 123 sequance as one item.)

so prob is 1/ ((N-1)*N)

anything else yu want to know ?

Answer 3

the same as any other randomly selected permutation...

Answer 4

formula in probability statistics