I think the no of combinations would be 4*3*5=60. so answer is 60.


@Leonswati: Requesting you not to post PS problem in DS section else the whole meaning of creating sections is lost.

should it not be 4.3.5/(3.2)

because for a single combination of ABC we can form 6 different combinations right ?

that makes it 10 but answer is not there ?
should we count ABC and BAC different or same set ?

leonswati wrote:Kendra is a faculty adviser at a college, and she has decided to help 4 student leaders, 3 faculty members, and 5 administrators meet and learn about each other. She decides to invite one student, one faculty member, and one administrator to have lunch with her every Friday. For how many Fridays can Kendra avoid having lunch with the exact same combination of student, faculty member, and administrator?
12
48
60
64
68


Plz help me solve this

Number of student leader options = 4.
Number of faculty member options = 3.
Number of administrator options = 5.
To combine the options above, we multiply:
4*3*5 = 60.

The correct answer is C.

When we're combining from DIFFERENT POOLS -- in this case, from a pool of students, a pool of faculty members, and a pool of administrators -- no division is necessary.

When we're combining from a SINGLE POOL, we need to divide by the number of ways to ARRANGE the elements being chosen.
In the problem above, the number of combinations of 3 that can be formed from the SINGLE POOL of 5 administrators = (5*4*3)/(3*2*1) = 10.