First method: Sort people into ten labeled groups and then recognize that you have overcounted. There are $\binom{50}{5,5,\dots,5}=\binom{50}{5}\binom{45}{5}\binom{40}{5}\cdots\binom{10}{5}=\frac{50!}{(5!)^{10}}$ ways that you can do this. Now, we recognize that each outcome we actually counted $10!$ different times. Dividing by $10!$ corrects this mistake giving us a final total of:
$$\frac{50!}{(5!)^{10}}\cdot\frac{1}{10!}$$
Second method: As "division by symmetry" arguments can make people nervous, we can instead approach more carefully to avoid overcounting. As we have 50 distinct people, there must be some obvious way that we can label them and arrange them. I will continue based on ordering the people by height, but you could choose a different ordering if you choose.
Among the 50 people, there must be someone who is shortest. We first pick who the four other people in the shortest person's group is. This can be accomplished in $\binom{49}{4}$ ways. We then remove all of these four people along with the original shortest person from the pool of available people to be placed in groups later.
Next, among the 45 remaining people, there must be someone who is shortest. We pick the four other people in the shortest remaining person's group. This can be accomplished in $\binom{44}{4}$ ways. Continue as before removing these people from the available pool.
Repeat this process until everyone has been assigned a group. This gives a final total number of ways as:
$$\binom{49}{4}\binom{44}{4}\cdots \binom{14}{4}\binom{9}{4}$$
One can check that this answer is equal to the previous answer.
Best Answer
A method to count this without needing to rely on dividing for symmetry:
Take your eight students and arrange them in some standard fashion. For the purpose of illustration, I will choose to do this by height, but other methods of ordering will work as well.
Take the shortest person in the group. (there is no choice to be made here, it is predetermined).
Pick a person not yet taken to be made the shortest person's group partner. (Seven choices available)
From those six people remaining, take the shortest remaining person in the group. (again, there is no choice to be made here, it is predetermined based on previous choices)
Pick a person still remaining to be made this person's group partner. (Five choices available)
$\vdots$ Repeat the process, taking the current shortest person from those remaining and then picking a partner for them.
Applying multiplication principle, there are $1\cdot 7\cdot 1\cdot 5\cdot 1\cdot 3\cdot 1\cdot 1 = 7\cdot 5\cdot 3 = 7!! = 105$ (here $7!!$ is double factorial notation)
Note that this is equal to $\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}\frac{1}{4!}=\binom{8}{2,2,2,2}\frac{1}{4!}=\frac{8!}{(2!)^44!} = 105$
Explaining the way the answer was written and how to view it that way,
Temporarily assign group names: group1, group2, group3, group4.
Arrange the eight people into these groups with 2 each in $\binom{8}{2,2,2,2}$ ways. (to see this, pick two to go to group1, then pick two remaining to go to group 2, etc... for a total of $\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$ ways)
Now, recognize that we have counted every scenario exactly $4!$ times each since each rearrangement of the group names yields the "same" outcome. Dividing by $4!$ corrects our count, making the total:
$\binom{8}{2,2,2,2}\frac{1}{4!}=105$