Choosing 2 shoes from 6 pairs of different shoes How to select 2 shoes from 6 pairs of shoes where in the selected shoes they are not from the same pair?
Why is this answer wrong?
$${6 \choose 1}{2 \choose 1}{5 \choose 1}{2 \choose 1}$$
My logic is first we choose a pair from six then choose one of the two shoes and do it again for the next one.
But the right answer in the text book was:
$${6 \choose 2}{2 \choose 1}{2 \choose 1}$$
 A: See your answer and the given answer have a very slight difference . Your answer has $\binom{6}{1}\binom{5}{1}=30$ while the given answer has $\binom{6}{2}=15$
Both of you are selecting two pairs. But in your case, arrangement is also taking place. For example, in your answer you're counting selecting $P_1$ first then $P_2$ and  selecting $P_2$ first then $P_1$ as different cases. Because you first choose a pair by $\binom61$ and then choose the other one by $\binom51$. So automatically you're doing arrangement also. But since arrangement is not required, we select two pairs at a single time by $\binom{6}{2}$ Everything else is all right
A: As Mark Bennet has commented, you have counted each set of two twice depending on which shoe is chosen first. To check the answer, here is an alternate method.
There are $\binom{12}{2}$ sets of $2$ that can be drawn and only $6$ drawn sets are from the same pair. Thus, there are
$$
\binom{12}{2}-6=60
$$
sets of two shoes that do not belong to the same pair.
A: There are $12$ choices for the first shoe, and then $10$ choices for the second shoe, but this counts permutations not combinations, so we need to group the selection of shoe A and shoe B with the selection of shoe B and shoe A together, hence we divide by $2$.
A: Had they been socks, the answer would just have been $\binom62=15$
but shoes have a $L$ and $R$,  thus $15*2^2 = 60$ ways
