Let S be the set of permutations of \((1,2,3,4,5,6)\) whose first term is not 1.

If we choose a permutation at random from S, what is the probability that the third term is equal to 3?


I am quite confused on how to start, so how do i start? If possible, can you please provide multiple hints, or the solution?

Max0815  Sep 16, 2018

let's first figure out how many permutations do have the 3rd term equal to 3


The first number we have 4 choices (not 1 or 3), the second number we have 4 choices, the 3rd number is 3, the 4th there are 3 choices, 5th 2 choices.


This is 4 x 4! = 96


The total number of permutations in S is just 5 x 5! = 600 (do you see why?)

and thus


P[term 3 = 3] = 96/600 = 4/25

Rom  Sep 16, 2018

Thank you very much :)

Max0815  Sep 16, 2018

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