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avatar+194 

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?

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

 Sep 16, 2018
 #1
avatar+6187 
+3

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

 Sep 16, 2018
 #2
avatar+194 
+1

Thank you very much :)

 Sep 16, 2018

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