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

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

Answer 2

`10C2\cdot41C2=(10!)/(2!8!)\cdot(41!)/(2!39!)=(9\cdot10)/(2)\cdot(40\cdot41)/(2)=36,900`