Thought process-

I think that theprobability of atleast one pair=(1 - probability of matching no pair)

so I proceeded as follows -

4 socks are drawn in 20 C 4 ways.

therefore n(s) = 20C4

Now let us say

Ist Row - 1 2 3 4 5 6 7 8 9 10

2nd Row 1' 2' 3' 4' 5' 6' 7' 8' 9' 10'

1st socks can be selected in 20 C 1 let say1from Ist row.

2nd sock must not be the 1' therefore it can selected from remaining 18 socks in

18 C 1 ways = 18 let say it is any socks like 6.

3rd sock must not be 1' as well as 6' therefore It can be drwn from the remaining 16 balls. let say it is 10.

similarly 4th ball is drawn in 14 ways

probability of not matching even a single pair is

1 - (20 . 18 . 16 . 14 )/20C4.

Please guide me If I am on right way or missing out something.

Also the book answer for this is 99/323.

Anshu.