Question

How many different committees can be formed from 10 teachers and 41 students if the committee consists of 2 teachers and 2 ​students?

Answer 1

$10C2\cdot41C2=\dfrac{10!}{2!8!}\cdot\dfrac{41!}{2!39!}=\dfrac{9\cdot10}{2}\cdot\dfrac{40\cdot41}{2}=36,900$

Answer 2

If we define "n choose r" as C(n,r)=n!/(r!(n-r)!)
where C(n,r) represents the number of ways (order not important) we can choose r objects out of n, then

Number of ways to choose teachers = 10 choose 2 = C(10,2), and
number of ways to choose students = 41 choose 2 = C(41,2)

So the number of different committees 
= C(10,2)*C(41,2)
= 45*820
= 36900