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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: $${6 \choose 2}{2 \choose 1}{2 \choose 1}$$

Hossein RM
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    The way I first thought about it was a bit different - there are 12 shoes and I choose the first one from $12$ options and the second from $10$ (the five undisturbed pairs) then divide by $2$ because each pair is counted twice depending on which shoe is chosen first. – Mark Bennet Mar 01 '23 at 08:00
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    It is always useful to do some simple checks. Your formula is 'exotic', you SHOULD calculate, you will find it gives 120. And the TOTAL number of possibilities is ${2} \choose{12} $ $=66$ ; 120 is greater than 66 : impossible. And the final result is 66 minus the number of possibilities to choose 2 shoes from the same pair, Result=$66-6=60$. – Lourrran Mar 01 '23 at 08:10

4 Answers4

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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

MathStackexchangeIsNotSoBad
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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.

robjohn
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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 $2$.

JMP
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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

true blue anil
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