This is the way I interpret the question:
Let's call the sock types A, B, C, D. The question is (I think) saying there are 20 A-type socks, 20 B-type socks, 20 C-type socks, 20 D-type socks. So we need to select 10 socks at random from all these socks (80 socks in all) mixed together. We need the probability of selecting 10 socks of just one type, then we need the separate probability of selecting 10 socks of just two different types. My solution is based on these assumptions.
PROBABILITY OF SELECTING JUST ONE TYPE
Let's select the first sock. Since there are 4 types, it doesn't matter which type we pick first. Let's say we pick an A-type sock. That leaves 79 socks. The next sock must be A-type. There are 19 A-types left so the probability of selecting an A-type again is 19/79.
The third sock must be another A-type. This time we only have 18 A-types and there are 78 socks left, so the probability of selecting another A-type is 18/78.
And so on. To get the final probability we need to find the product of all the individual probabilities:
(19/79)(18/78)(17/77)(16/76)(15/75)(14/74)(13/73)(12/72)(11/71)=0.00000044884.
PROBABILITY OF SELECTING JUST TWO TYPES
Again, it doesn't matter which type the first sock selected is. This time it doesn't matter what type the second sock is. But the third sock and all the remaining socks have to be one of the two types already picked.
There are 78 socks left and out of these there are 19 of each of the two types already selected. That's 38 socks. The question doesn't say how many of the two different types we must end up with. So we could have 7 of one type and 3 of the other, or 4 of one and 6 of the other. If that's the case then we can work out the combined probability:
(38/78)(37/77)(36/76)(35/75)(34/74)(33/73)(32/72)(31/71)=0.0020857
Now, it's quite possible that the question has been wrongly interpreted. If so, perhaps you can clarify the question.
