Number Of Shoes

No two shoes can come from the same pair, so we must select four different pairs to choose shoes from. There are $5$ ways to do this (each choice of four pairs leaves out one pair).

Now, we have two choices of shoe from each of these four pairs. That means there are $2^4$ ways to select one shoe from each pair.

In total, we have $5\times 2^4 = 5\times 16 = \boxed{80}$ ways to chose four shoes from different pairs.

Let $A_j$ (for $1 \le j \le 5$ ) be the set of choices of four shoes that contain the $j$ th pair of shoes. Then $\begin{aligned} |A_j| & = \; \binom{8}{2} = 28 & & 1 \le j \le 5 \\ |A_j \cap A_k| & = \; 1 & & 1 \le j < k \le 5 \\ A_i \cap A_j \cap A_k & = \; \varnothing & & 1 <i < j < k \le 5 \end{aligned}$ so that $|A_1 \cup A_2 \cup A_3 \cup A_4 \cup A_5| \; = \; \sum_{j=1}^5 |A_j| - \sum_{j < k}|A_j \cap A_k| \; = \; 5 \times 28 - 10 \times 1 = 130$ and hence the number of ways of choosing four shoes so that no pair has been chosen is $\binom{10}{4} - 130 \; = \; \boxed{80}$