# Number of ways to choose shoes so that there will be no complete pair

I am just stuck with a problem and need your help. The question is:

A closet has 5 pairs of shoes. The number of ways in which 4 shoes can be chosen from it so that there will be no complete pair is a.80 b.160 c.200 d.none

I think the answer should be $\binom{10}{1}\cdot\binom{8}{1}\cdot\binom{6}{1}\cdot\binom{4}{1}$. But I seriously doubt this could be the answer. Please help me in solving this.

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For a specific set of 4 pairs, there are $2^4$ ways of choosing the shoes. There are 5 possible sets of 4 pairs. Hence there are $5*2^4$ ways in total.
Actually your idea is right, but you are counting the number of ordered sequences of four shoes. Divide by $4!$ and you get the answer you are looking for.