The answer depends on whether or not the tickets are numbered.
Suppose that each ticket simply says: "Admit two people, as long as they are a male/female pair" (backward, and illegal where I live). We can proceed symmetrically as follows.
The females can be chosen in $\binom{7}{4}$ ways. For each choice of females, the males can be chosen in $\binom{6}{4}$ ways, for a total of $\binom{7}{4}\binom{6}{4}$ ways.
Now we need to do the pairing. The females line up in order of (say) height, or student number. The first female picks her partner. She has $4$ choices. Then the next female chooses. For every way that the first female chose, there are $3$ ways for the second female to choose, and so on. So once the people have been chosen, for each choice of people there are $4!$ ways to do the pairing, for a total of $\binom{7}{3}\binom{6}{4}(4!).$
An alternate way to do the problem is to note that the females can be chosen in $\binom{7}{4}$ ways. Line them up in order of height, and let the females, in order, make their choices. The tallest female has $6$ choices, for each such choice the second tallest female has $5$, and so on, for a total of $\binom{7}{4}(6)(5)(4)(3).$
Enumerating is unpleasant. Give the females names, such as A, B, C, D, E, F, G, H and also the males, say U, V, W, X, Y, Z. A sensible way of listing the choices is to list the females who were chosen in alphabetic order, with each female name followed by the name of the male she is partnered with. For example, the "word" BZDUEXGY indicates that B is partnered with Z, D with U, and so on. Now enumerate all $8$-letter words of the right shape, in lexicographic (dictionary) order.
Remark: If the tickets are numbered, we just multiply the answer above by $4!$. Alternately, line up the tickets in ticket number order. For the first ticket, there are $(7)(6)$ ways to choose the couple that will get it. For each such choice, there are $(6)(5)$ ways to choose the couple that will get the second ticket, and so on, for a total of $(7)(6)(6)(5)(5)(4)(4)(3).\qquad$ From this, if we wish, we can derive the answer for unnumbered tickets. For every unnumbered ticket possibility, there are $4!$ numbered ticket possibilities, so we simply divide the above expression by $4!$.