# 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.

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.

• Why should I divide by 4!? – Mistu4u Oct 1 '12 at 4:36

We need to chose $$4$$ different color shoes from the five different pairs of shoes given. Therefore number of ways to select them is $$\displaystyle \binom{5}{4}$$. Now from a pair of two shoes we need to select only one shoe so that we don't get a complete pair of shoes in our four selected $$4$$ shoes. The number of ways to select them is $$\displaystyle \binom{2}{1}$$. We need to do this with $$4$$ different times Therefore their number of ways becomes $$\displaystyle\binom{2}{1}^4$$. Therefore the total number of ways to choose $$4$$ shoes will be $$\displaystyle \binom{5}{4}\cdot\binom{2}{1}^4$$ which is equal to $$80$$.

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 $4$.