+198 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?
+6250 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
