Partitions?

Question

Hi.
Could someone please walk me through the solution of this problem and also explain how you arrive at the solution:

In how many ways can 10 people be divided into 3 teams, one containing 4 and the other 3.
Thank you

Answer 1

First think of the first team with 4.
That would be (10*9*8*7)/(4!).
Then the next team with 3.
Thats (6*5*4)/(3!).
Now since there would be only 1 way to place the last team with last 3,
then its ((10*9*8*7)/(4!))*((6*5*4) /(3!)) = 1050 ways I think.

Answer 2

10C4 = 10!/(4!*6!) possible teams by choosing 4 people at random.

6C3 = 6!/(3!*3!) possible teams of 3, choosing from the remaining folks.

There is only one way to pick the last team of three--the three people not yet chosen.

10C4 + 6C3 + 1 = possible number of teams having 4, 3, and 3 members.