6. Dragomir has 6 pairs of socks in a drawer. Each pair of socks is different from every other pair in the drawer. He randomly removes 4 individual socks from the drawer, one at a time. What is the probability that there is exactly 1 matching pair of socks among these 4 socks?
6. Solution 1
Suppose that Dragomir removes the socks one at a time. There are 12 socks that he can remove 1st, then 11 socks 2 nd, then 10 socks 3 rd, and finally 9 socks 4 th.
This means that there are $12 \times 11 \times 10 \times 9$ ways in which he can remove 4 socks one at a time. Suppose that there is exactly one pair among the 4 socks removed. There are 6 different possibilities for how this pair was formed: 1st and 2nd socks match, 1st and 3rd match, 1st and 4th match, 2nd and 3rd match, 2nd and 4th match, 3rd and 4th match.
We determine the number of ways in which each of these possibilities can arise.
Case 1: Matching pair 1st and 2nd
There are 12 socks that he can pick 1 st (no restrictions).
There is 1 sock that he can pick 2nd (pair of 1st).
There are 10 socks that he can pick 3rd (any of the 10 remaining socks).
There are 8 socks that he can pick 4th (any of the 9 remaining socks other than pair of 3rd). In this case, there are $12 \times 1 \times 10 \times 8$ ways.
Case 2: Matching pair 1st and 3rd
There are 12 socks that he can pick 1st (no restrictions).
There are 10 socks that he can pick 2nd (all but pair of 1st).
There is 1 sock that he can pick 3rd (pair of 1st).
There are 8 socks that he can pick 4th (any other than pair of 2 nd).
In this case, there are $12 \times 10 \times 1 \times 8$ ways.
Case 3: Matching pair 1st and 4th
There are 12 socks that he can pick 1st (no restrictions).
There are 10 socks that he can pick 2nd (all but pair of 1st).
There are 8 socks that he can pick 3rd (all but pair of 1st and pair of 2 nd).
There is 1 sock that he can pick 4th (pair of 1st).
In this case, there are $12 \times 10 \times 8 \times 1$ ways.
Case 4: Matching pair 2nd and 3rd
There are 12 socks that he can pick 1 st (no restrictions).
There are 10 socks that he can pick 2nd (all but pair of 1st).
There is 1 sock that he can pick 3rd (pair of 2nd).
There are 8 socks that he can pick 4 th (all but pair of 1st).
In this case, there are $12 \times 10 \times 1 \times 8$ ways.
Case 5: Matching pair 2nd and 4th
There are 12 socks that he can pick 1st (no restrictions).
There are 10 socks that he can pick 2nd (all but pair of 1st).
There are 8 socks that he can pick 3rd (all but pair of 1st and pair of 2nd).
There is 1 sock that he can pick 4th (pair of 2 nd).
In this case, there are $12 \times 10 \times 8 \times 1$ ways.
Case 6: Matching pair 3rd and 4th
There are 12 socks that he can pick 1st (no restrictions).
There are 10 socks that he can pick 2nd (all but pair of 1st).
There are 8 socks that he can pick 3rd (all but pair of 1st and pair of 2 nd).
There is 1 sock that he can pick 4th (pair of 3rd).
In this case, there are $12 \times 10 \times 8 \times 1$ ways.
In each of the six cases, there are $12 \times 10 \times 8 \times 1$ ways in which the socks can be chosen.

Therefore, the overall probability is $\frac{6 \times 12 \times 10 \times 8 \times 1}{12 \times 11 \times 10 \times 9}=\frac{6 \times 8}{11 \times 9}=\frac{2 \times 8}{11 \times 3}=\frac{16}{33}$.
Solution 2
Since there are 12 socks in the drawer, there are $\binom{12}{4}=\frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1}=11 \times 5 \times 9=495$ ways of choosing 4 socks.
Next, we count the number of ways in which the 4 chosen socks can include exactly 1 matching pair.
There are 6 possible matching pairs that can be chosen.
The remaining 2 socks must come from 2 different pairs of the remaining 5 pairs. If we label the remaining pairs A, B, C, D, E, there are 10 possible two-pair combinations from which these 2 socks can be chosen: $\mathrm{AB}, \mathrm{AC}, \mathrm{AD}, \mathrm{AE}, \mathrm{BC}, \mathrm{BD}, \mathrm{BE}, \mathrm{CD}, \mathrm{CE}, \mathrm{DE}$. (We could also use a binomial coefficient to calculate this total.)
For each of the 2 pairs from which these 2 socks will be chosen, there are in fact 2 socks to choose
Thus, the total number of ways in which the 4 socks can be chosen that satisfy the given conditions is $6 \times 10 \times 2 \times 2=240$.
This means that the probability is $\frac{240}{495}=\frac{16 \times 15}{33 \times 15}=\frac{16}{33}$.
Solution 3
Since there are 12 socks in the drawer, there are $\binom{12}{4}=\frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1}=11 \times 5 \times 9=495$ ways of choosing 4 socks.
These 4 socks can include 0 matching pairs, 1 matching pair, or 2 matching pairs. We count the number of ways in which the 4 chosen socks can include 0 matching pairs or 2 matching pairs and subtract this from the total number of ways of choosing 4 socks to obtain the number of ways of choosing 1 matching pair.
To choose 0 matching pairs, we choose 4 of the 6 pairs and choose 1 of the 2 socks in each pair. There are $\binom{6}{4}=\frac{6 \times 5 \times 4 \times 3}{4 \times 3 \times 2 \times 1}=15$ ways of choosing 4 of 6 pairs, and $2^{4}=16$ ways of choosing 1 of the 2 socks in each of 4 pairs.
Thus, there are $15 \times 16=240$ ways of choosing exactly 0 matching pairs.
To choose 2 matching pairs, we choose 2 of the 6 pairs and choose both socks in each pair.
There are $\binom{6}{2}=\frac{6 \times 5}{2 \times 1}=15$ ways of choosing 2 of 6 pairs, and 1 way to choose both socks in each pair.
Thus, there are $15 \times 1=15$ ways of choosing exactly 2 matching pairs.
This means that there are $495-240-15=240$ ways of choosing exactly 1 matching pair. This means that the probability is $\frac{240}{495}=\frac{16 \times 15}{33 \times 15}=\frac{16}{33}$.
ANSWER: $\frac{16}{33}$