There are only 10c4 = 2 10 ways to choose 4 from 10 different shoes, among which 5 C4 * 2c 1 * 2c 1 = 80 ways to choose 4, so it meets the requirements.

Method 2, (direct calculation)

From the five pairs of shoes, there are 5c 1 pairs, of which 5C 1=5Z, and then from the remaining eight pairs of shoes, there are 8C2=28 pairs, and * * * there are 5C 1*8C2= 140 pairs, including the first pair, the second pair and the first pair.

Method 3, (direct calculation)

Divided into two categories:

The first category: four shoes are just made into two pairs by 5C2= 10 method;

The second type: there are two pairs, and the other two pairs are not, using 5c1(8c1* 6c1)/2 =120 method;

* * * There are 130 methods.