Committees Formed--8.2

Question

How many different committees can be formed from 11 teachers and 49 students if the committee consists of 3 teachers and 2 students?

Answer 1

Hello, Amanda

 

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.  Please rate my answer.  I am assuming that you are doing combinations and permutations. 

 

D. Y. Taylor