A professor will assign seven projects to three students. If

I solved the first problem like this:

7C2 * 5C2 * 3C3
21 * 10 * 1
210

Since there are 3 ways any one student can get 3 projects while the other 2 get 2 each (3C1), multiply the previous answer by 3 for the final answer.

210 * 3 = 630

This is different from the given answer though, so I may be missing something.

I see... I hadn't considered accounting for the variability of which student would get additional projects.

It's apparent, however, that the questions as posed are not in their original wording. I wonder if the original wording is clearer regarding this variability?

Edit:

Also, you could make the same argument for question two: that the variability of which project gets additional students would have to be accounted for.

I think the 610 should be 210, the questions are indeed not only 'similar' but structurally identical.

Both the problems should have the same answer. I get 630 for both the problems.
There are 3 ways to choose which group should have 3 projects assigned.
There are 7c2 * 5c2 *3c3 ways to assign them.

So 3 * 7c2 * 5c2 *3c3 = 630.

The projects are all distinct and so are the students. So in both the cases, you are doing essentially the same. In the first case, you are assigning 7 things to 3 people and in the second case, you are assigning 7 people to 3 things.

So in both the cases, you choose the person/thing that will get 3 things/person in 3C1 = 3 ways.
Next, you choose 3 things/people out of 7 in 7C3 ways.
Next, you choose 2 things/people out of the remaining 4 in 4C2 ways.

3C1 * 7C3 * 4C2 = 3*(7*6*5)/(3*2) * 6 = 630 ways

Responding to a pm:



This is the reason your logic doesn't work:

The problem is with distributing 4 projects between 2 students such that each student gets 2 projects each.

Say students are Sa and Sb
Say projects are P1, P2, P3 and P4.

4C2 is the way you select 2 projects for Sa. Whatever is left is for Sb.

Say, you select
P1 and P2 for Sa. You have P3 and P4 for Sb.
P1 and P3 for Sa. You have P2 and P4 for Sb.
P1 and P4 for Sa. You have P2 and P3 for Sb.
P2 and P3 for Sa. You have P1 and P4 for Sb.
P2 and P4 for Sa. You have P1 and P3 for Sb.
P3 and P4 for Sa. You have P1 and P2 for Sb.

There are 6 ways which is 4C2.

Look at what happens when you do 4C2*2C1.
You select P1 and P2 for Sa. You have P3 and P4 for Sb. Then since you multiply by 2, you select P1 and P2 for Sb. You have P3 and P4 for Sa.
In another case, you select P3 and P4 for Sa. You have P1 and P2 for Sb. Then since you multiply by 2, you select P3 and P4 for Sb. You have P1 and P2 for Sa.

Note that you have double counted both cases. When you select 2 out of 4, you are selecting them for one particular person. So you do not multiply by 2C1.

I think it's 1260. Here's my answer:
2 projects out of total of 7 can be selected in 7C2=21 different ways. These two project is given to 1 out of the 3 students in 3C1=3 different ways.
This can be done in 63 different ways.
Again, out of remaining 5 projects, 2 is to be selected in 5C2 = 10 different ways. These 2 projects is assigned to 1 out of remaining 2 students in 2C1=2 ways.
This can be done in 20 different ways.
Now, at last, 3 projects are remaining out of which 3 is selected in 3C3=1 way and assigned to remaining 1 student.
This can be done in only 1 way.
Thus, total no. of ways in which the process can be done is 63*20*1=1260 ways.