How many different committees can be formed from 11 teachers and 44 students if the committee consists of 3 teachers and 2 students?

*this question is in my permutations and combinations section and this is the full question

3 of because 3*3=9 and we can't have more than that

 From a group of 11 teachers, we need to select 3 teachers: C(11,3) (11!)/[(8!)(3!)]= 11*10*9*8! =990= 165          Note: 8! will cancel each other out                          8!3*2*1         6 From a group of 49 students, we need to select 2 students: C(49, 2) (49!)/[(47!2!)]= 49*48*47! = (49)(48)= 2352= 1176                            47!*2*1             2          2 Again, 47! will cancel 47! Now, multiply the two answers: (165)(1176)= 194,040 different committees can be formed.  I hope this helps.