Seating people in $2$ rows with conditions

Question

$10$ students are to be seated in $10$ seats. The seats are in $2$ rows of $5$ seats each. $3$ students are nearsighted and need to sit in the front. $2$ students are farsighted and must sit in the back. How many ways are there to seat the students?

Answer 1

The setup to the answer:

First we need to choose $3$ of the $5$ front seats for the three nearsighted students. Then we can arrange the nearsighted students in the chosen seats in $3!$ ways.

Second, we must choose $2$ of the $5$ back seats for the farsighted students. Then we can arrange the farsighted students in the chosen seats in $2!$ ways.

Finally we must sit down the remaining $5$ remaining students in the remaining chairs.

Now see if you can use this to figure out the problem

Answer 2

Select the seats for the near sighted people in $\binom{5}{3}$ ways and permute internally in $3!$ ways.

Select the seats for the far sighted people in $\binom{5}{2}$ and permute internally in $2!$ ways.

Assign the remaining seats in $5!$ ways.

Final answer is $\binom{5}{3}\binom{5}{2}5!3!2!$