Combination and Probability There are n students and n+2 different gifts.  Each student have to receive 1 gift package. How many ways can we give out all the gifts.
Scenario 1. A gift package has 3 gifts.

Ways to choose 3 gifts: (n+2)Choose(3)


Scenario 2: 2 gift packages has 2 gifts in it.

Dont know how to do..

Finally: These gift packages can be permutated and given

(Scenario 1 + Scenario 2) * n!

 A: It looks as if you have a problem only with the case that two gift packages have two gifts. The $4$ gifts in these packages can be chosen in $\binom{n+2}{4}$ ways. For every such way, there are $3$ ways to split the group of $4$ into two groups of $2$. For the "nicest" gift among the $4$ can be put into a package with any of the three others.
That gives a total of $3\binom{n+2}{4}$ for your Scenario 2.
Another way: Alternately, imagine first that two of the gifts will be put in a blue box, and two in a red box. This can be done in $\binom{n+2}{2}\binom{n}{2}$ ways. However, this double-counts the number of ways to split the gifts for Scenario 2. So Scenario 2 can happen in $\frac{1}{2}\binom{n+2}{2}\binom{n}{2}$ ways.
A: First case:  First choose the three gifts that go in the one bonanza gift package ($_{n+2}C_3$) then choose which student gets it ($n$).  Distribute the remaining $n-1$ gifts to the other students, one apiece, $(n-1)!$ ways.
Second case:  Chose the four gifts that go into the two bonanza gift packages $(_{n+2}C_4)$ then choose how you divide them up ($3$).  Choose the two students to get the bonanza gift packages $(_nC_2)$ and then choose which student gets which package ($2$).  Distribute the remaining $n-2$ gifts to the other students, one apiece, $(n-2)!$ ways. 
