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Suppose we uniformly and randomly select a permutation from the 20! permutations of 1,2,3,...,20. What is the probability that 2 appears at an earlier position than any other even number in the selected permutation?

1. 1/2
2. 1/10
3. 9!/20!
4. None of the above.

Correct answer: 1/10

Solution

Among the 10 even numbers {2,4,...,20}, by symmetry each is equally likely to appear first within that subset. So the probability that 2 precedes all other evens is 1/10, not 1/2.

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