Samuel has 5 different pairs of socks for each week. Each sock in a pair is indistinguishable from one another (no left or right socks). Samuel doesn’t care how crusty he looks and randomly chooses two socks to wear for each weekday (on Friday, he will only have 2 socks to choose from and must wear them). At the end of each day, he throws the two socks he wore that day in his hamper (even though he is crusty, he does not wear the same socks again until they are washed at the end of the week on Saturday). If the probability that Samuel never has a correct pair of socks for any of the 5 weekdays can be written as , where and are coprime positive integers, find .
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Samuel takes the 1 0 socks in some order. He wears the first two on Monday, the second two on Tuesday, and so on.
There are 2 5 1 0 ! = 1 1 3 4 0 0 orders in which the 1 0 socks can be chosen, all equally likely.
Let A j be the set of orderings for which Samuel wears matching socks on day j , for 1 ≤ j ≤ 5 .
Using the Inclusion-Exclusion Principle, we deduce that ∣ ∣ A 1 ∪ A 2 ∪ A 3 ∪ A 4 ∪ A 5 ∣ ∣ = 5 × 1 2 6 0 0 − 1 0 × 1 8 0 0 + 1 0 × 3 6 0 − 5 × 1 2 0 + 1 2 0 = 4 8 1 2 0 Thus the probability that Samuel never wears a pair of matching socks is 1 1 3 4 0 0 1 1 3 4 0 0 − 4 8 1 2 0 = 9 4 5 5 4 4 making the answer 5 4 4 + 9 4 5 = 1 4 8 9 .