# 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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Hint: The four shoes must come from 4 different pairs.

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.

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Why should I divide by 4!? –  Mistu4u Oct 1 '12 at 4:36
Since we have $5$ pairs of shoes, and we wish to choose $4$ shoes avoiding pairs of shoes, the answer is the number of independent $4$-sets of vertices of the graph:
I.e., it is the coefficient of $x^4$ in its independence polynomial.
We know that the independence polynomial of a disconnected graph is equal to the product of independence polynomials of its connected components. The independence polynomial of $K_2$ is $1+2x$, so the independence polynomial of the above graph is $$(1+2x)^5=32x^5+80x^4+80x^3+40x^2+10x+1.$$ Hence there are $80$ independent sets of size $4$.