In how many ways can $4$ different packages be distributed among $8$ people such that everyone gets $2$ packages or none? I need some helps in terms of explanations for the answer for an exercise I’ve got in my lecture 
The question is how many ways you can distribute $4$ different packages among $8$ people A, B,..., H, such that everyone gets $2$ packages or none.
The professor has given the answer, which is $\binom{8}{2}\binom{4}{2} = 28 \cdot 6$.
$\binom{4}{2}$, which is fine since we have to choose $2$ out of $4$ packages for distribution.
The part I can’t understand is why we have to choose specifically $2$ people but not $1$? The question does not imply that $2$ people should be chosen but instead, stated that, ”everyone.”
Thanks for your help in advance :):) 
 A: In this context, "everyone" means "each one".

Since there are $4$ packages, and each person gets either $2$ packages or none, it follows that exactly $2$ people will get packages, each of them getting $2$ packages.

To count the number of ways to distribute the packages . . .

First choose the $2$ people who will be given packages: $\binom{8}{2}$ choices.

Next choose $2$ packages to give to the first (order them arbitrarily) of the two chosen people:  $\binom{4}{2}$ choices.

Give the remaining two packages to the second person.

With those choices, the distribution is complete.

By the multiplication rule, the number of possible distributions is $\binom{8}{2}\binom{4}{2}$.
A: You need to pick $2$ people because you have $4$ packages, and each person can only receive $0$ or $2$ packages.
A: Think like this:
Finally the four packages will be distributed, and since one person take two or none, there must be two lucky guys out of eight, so it's:
$$\binom{8}{2}.$$
And since we haven't decide that for each guy, which two types they receive? So let's choose for the first guy, and let the remainder for the second:
$$\binom{4}{2}.$$
