Choosing 4 people from 5 pairs My question is this:

$4$ students are chosen at random from $5$ pairs of $2$ students (i.e. $10$ students). What is the probability that no students are chosen from the same pair (each of the 4 chosen students is selected from a different group of two)?

I am very stuck on this question. I think it is $10C4$ - ways that both students of one group are chosen - ways that both students of $2$ groups are chosen, but I cannot figure out how to find the probability that both students of one group are chosen.
Any guidance would be helpful.
 A: Method 1: Ordered selections.
There are $10 \cdot 9 \cdot 8 \cdot 7$ ways to make an ordered selection of four of the ten students.
There are $10 \cdot 8 \cdot 6 \cdot 4$ ways to make an ordered selection of four students from different pairs.
Hence, the probability that no two students from the same pair are selected is
$$\frac{10 \cdot 8 \cdot 6 \cdot 4}{10 \cdot 9 \cdot 8 \cdot 7}$$
Method 2: Unordered selections.
There are $\binom{10}{4}$ ways to select four of the ten students.
There are $\binom{5}{4}$ ways to select four of the five pairs from which to select a student.  For each such pair, there are two ways to select a student from that pair.  Hence, the number of ways to select a student from four different pairs is
$$\binom{5}{4}2^4$$
Hence, the probability that no students from the same pair are selected is
$$\frac{\dbinom{5}{4}2^4}{\dbinom{10}{4}}$$
Method 3: Complementary counting.
As mentioned above, there are
$$\binom{10}{4}$$
ways to select four of the ten people.
We count selections in which both members of at least one pair of people are selected.
Both members of a pair are selected:  There are five ways to select the pair from which both members are selected and $\binom{8}{2}$ ways to select two of the other eight people.  Hence, there are
$$\binom{5}{1}\binom{8}{2}$$
such selections.
However, if we subtract this amount from the total, we will have subtracted too much since we will have subtracted each selection in which both members of two pairs are selected twice, once for each way we could have designated one of those two pairs as the pair from which both members were selected.  We only want to subtract such cases once, so we must add the cases in which both members of two pairs are selected to our total.
Both members of two pairs are selected:  There are
$$\binom{5}{2}$$
ways to select the two pairs from which both members are selected.
Hence, by the Inclusion-Exclusion Principle the number of favorable cases is
$$\binom{10}{4} - \binom{5}{1}\binom{8}{2} + \binom{5}{2}$$
Therefore, the probability that no two students from the same pair are selected is
$$\frac{\dbinom{10}{4} - \dbinom{5}{1}\dbinom{8}{2} + \dbinom{5}{2}}{\dbinom{10}{4}} = 1 - \frac{\dbinom{5}{1}\dbinom{8}{2} - \dbinom{5}{2}}{\dbinom{10}{4}}$$
