happyface101 wrote:
Can someone please help me understand why is 6C3 (choose 3 students from 6 students) = 6!/(3!3!) so 6*5*4/(3*2) -->20 ways, instead of simply 6*5*4 following the reasoning that choice #1 - you can pick from 6 students, choice #2 - you can pick from 5 students, and finally choice #3 - you can pick from 4 students.
Thanks so much!
Hi,
See the Question,
AT LEAST 1 Professor In every scenario, you need to count at least 1 or more than 1 professor, to obtain 3 Supervisory committee member.
Case 1. :- 1 Professor, 1 student, 1 Student = 3 Total member.
Now, using the slot method or directly using combination formula, you can select 1 Professor out of 4 {4c1}, in
4 way. Similarly you can select 2 student out of 6 by {6c2} =
15 way .
Now, as per Multiplication Principle, 15 * 4 = 60.
Case 2. :- 2 Professor, 1 student = 3 total member.
Using the Combination formula, 2 Professor out of 4 {4c2}, in
6 way. Similarly for 1 student out of 6 by {6c1} =
6 way .
-> 6 * 6 = 36.
Case 3:- Similarly we can go for 3 professor (remember at least one and no limitation till three, as only 3 Supervisory committee member required), without selecting any student.
This can done in {6c3} =
4 way.
Add, case 1 + Case 2 + case3 -> 60 + 36 +4 = 100, is the correct answer.