In how many ways can 10 students be divided into three teams, one containing 4 students and the others 3?

Method 1 makes sense, in which the answer is $\displaystyle \frac{10!}{4!3!3!}\cdot\frac{1}{2}$.

But method 2 bothers me, where the answer is $\displaystyle \binom{10}{4}\binom{5}{2}$.

I can understand choosing 4 out of 10 the first time, but what is the logical explanation for choosing 2 out 5 the second time? This does not seem very consistent.