probability problem/combinations Confused on how to do this.
4 people play a game of cards. These cards are marked with three out of five symbols (A,B,C,D, and E). Each person gets to create one custom card of their preference with any three symbols they like. What is the probability that exactly 2 people from the 4 players end up having the same custom card with the same symbols? (Assume you cannot repeat the same symbol on a card, for instance, you cannot have a card with the symbol combination ACA, DBB, or EEE).
My attempt:
$$P(X=2) = \frac{10 \times10 \times \dbinom{4}{2}^2}{\dbinom{5}{3}^4} = 0.06$$
Denominator because multiplication rule of how many ways 4 people can choose a custom card (3 unique symbols chosen from 5). Numerator: If two people have the same custom card, there are 6 ways to do it hence 4 choose 2, and then the remaining possibilities for the other 2 people is 10 times 10. Is my logic/understanding here ok?
 A: Not quite.  Since exactly two of the people must have the same symbols on their card, there are not $\binom{5}{3} =  10$ possible choices for the symbols on the cards of the remaining people since they must be distinct from the choice made by the people who chose the same three symbols and from each other.
There are $\binom{4}{2}$ ways to select the two people who have the same set of symbols on their card, $\binom{5}{3}$ ways to choose which three of the five symbols appear on the cards of the people who have the same symbols on their card, $10 - 1 = 9$ ways to select the symbols which appear on the card of the younger of the two remaining people, and $9 - 1 = 8$ ways to pick the symbols on the card of the remaining person.  Hence, the number of favorable cases is
$$\binom{4}{2}\binom{5}{3} \cdot 9 \cdot 8$$
Therefore, the probability that exactly two of the four people will select the same three symbols for their custom cards is
$$\Pr(X = 2) = \frac{\dbinom{4}{2}\dbinom{5}{3} \cdot 9 \cdot 8}{\dbinom{5}{3}^4}$$
