Probability Problem | Combinatorics | AIME I, 2015 - Question 5

In a drawer Sandy has 5 pairs of socks, each pair a different color. on monday sandy selects two individual socks at random from the 10 socks in the drawer. On tuesday Sandy selects 2 of the remaining 8 socks at random and on wednesday two of the remaining 6 socks at random. The probability that wednesday is the first day Sandy selects matching socks is \(\frac{m}{n}\), where m and n are relatively prime positive integers, find m+n.

Wednesday case - with restriction , select the pair on wednesday in \(5 \choose 1 \) ways

Tuesday case - four pair of socks out of which a pair on tuesday where a pair is not allowed where 4 pairs are left,the number of ways in which this can be done is \(8 \choose 2\) - 4

Monday case - a total of 6 socks and a pair not picked \(6 \choose 2\) -2

by multiplication and principle of combinatorics \(\frac{(5)({5\choose 2} -4)({6 \choose 2}-2)}{{10 \choose 2}{8 \choose 2}{6 \choose 2}}\)=\(\frac{26}{315}\). That is 341.

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