Suppose we uniformly and randomly select a permutation from the 20!…
2007
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?
- A.
1/2
- B.
1/10
- C.
9!/20!
- D.
Node of the above
Attempted by 72 students.
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Correct answer: B
Answer: 1/10
Reason (symmetry): There are 10 even numbers (2, 4, 6, ..., 20). In a uniformly random permutation all relative orders of these 10 even numbers are equally likely, so each even number is equally likely to be the earliest among the evens. Therefore the probability that 2 is earliest among the even numbers is 1/10.
Reason (counting):
Choose which 10 positions out of 20 will be occupied by even numbers: (20 choose 10) ways.
Place 2 in the earliest of those chosen positions, arrange the remaining 9 even numbers in 9! ways, and arrange the 10 odd numbers in 10! ways.
Favorable permutations = (20 choose 10) * 9! * 10!. Total permutations = 20!. Dividing gives ((20 choose 10) * 9! * 10!) / 20! = 1/10.
Hence the probability that 2 appears earlier than any other even number is 1/10.