Pairing Socks

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 a b \frac{a}{b} , where a a and b b are coprime positive integers, find a + b a+b .


The answer is 1489.

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1 solution

Mark Hennings
Jul 25, 2017

Samuel takes the 10 10 socks in some order. He wears the first two on Monday, the second two on Tuesday, and so on.

There are 10 ! 2 5 = 113400 \frac{10!}{2^5} = 113400 orders in which the 10 10 socks can be chosen, all equally likely.

Let A j A_j be the set of orderings for which Samuel wears matching socks on day j j , for 1 j 5 1 \le j \le 5 .

  • for any 1 i 5 1 \le i \le 5 , if Samuel wears matching socks on day i i , then there are 5 5 choices of which pair of socks he will wear that day, and then 8 ! 2 4 \frac{8!}{2^4} ways of choosing socks for the other four days. Thus A i = 5 × 8 ! 2 4 = 12600 |A_i| \; = \; 5 \times \frac{8!}{2^4} \; = \; 12600
  • for any 1 i < j 5 1 \le i < j \le 5 , if Samuel wears matching socks on days i i and j j ,then there are 5 5 choices of colour for the socks on day i i , then 4 4 choices of colour for the socks on day j j , and then 6 ! 2 3 \frac{6!}{2^3} ways of choosing the socks on the other three days. Thus A i A j = 5 × 4 × 6 ! 2 3 = 1800 |A_i \cap A_j| \; = \; 5 \times 4 \times \frac{6!}{2^3} \; =\;1800
  • Similarly, for any 1 i < j < k 5 1 \le i < j < k \le 5 , A i A j A k = 5 × 4 × 3 × 4 ! 2 2 = 360 |A_i \cap A_j \cap A_k| \; = \; 5 \times 4 \times 3 \times \frac{4!}{2^2} \; = \; 360
  • Similarly, for any 1 i < j < k < m 5 1 \le i < j < k < m \le 5 , A i A j A k A m = 5 × 4 × 3 × 2 × 2 ! 2 = 120 |A_i \cap A_j \cap A_k \cap A_m| \; =\; 5 \times 4 \times 3 \times 2 \times \frac{2!}{2} \; = \; 120
  • Finally A 1 A 2 A 3 A 4 A 5 = 5 × 4 × 3 × 2 × 1 = 120 |A_1 \cap A_2 \cap A_3 \cap A_4 \cap A_5| \; = \; 5 \times 4 \times 3 \times 2 \times 1 \; = \; 120

Using the Inclusion-Exclusion Principle, we deduce that A 1 A 2 A 3 A 4 A 5 = 5 × 12600 10 × 1800 + 10 × 360 5 × 120 + 120 = 48120 \begin{aligned} \big|A_1 \cup A_2 \cup A_3 \cup A_4 \cup A_5\big| & = 5 \times 12600 - 10\times1800 + 10\times360 - 5\times120 + 120 \\ & = 48120 \end{aligned} Thus the probability that Samuel never wears a pair of matching socks is 113400 48120 113400 = 544 945 \frac{113400 - 48120}{113400} \; = \; \frac{544}{945} making the answer 544 + 945 = 1489 544 + 945 = \boxed{1489} .

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