From five faculty members and four students, a committee of four is to be chosen that includes two students and two faculty members. In how many ways can this be done?
Please show more than one approach.
I would put in 9 nCr 4 in a calculator. Am I right?
From five faculty members and four students, a committee of four is to be chosen that includes two students and two faculty members. In how many ways can this be done?
Is it combinations problem? I think it is, so I would put 9 nCr 4.
Yeah, but two committee members must be chosen from faculty and two from students, then you have to start by separating the two groups and working from there.
By the way, if I find out who you are, you get a zero for the project.
Happy Monday!
To solve this problem, we can use both combinations and permutations.
Approach 1: Using combinations
We need to choose 2 students out of 4 and 2 faculty members out of 5. We can calculate this using combinations.
Number of ways to choose 2 students out of 4= C(4, 2)
Number of ways to choose 2 faculty members out of 5 = C(5, 2)
To calculate C(n, r), we use the formula: C(n, r) = n! / (r!(n-r)!)
Therefore, the number of ways to choose the committee using combinations = C(4, 2) * C(5, 2) = (4! / (2!(4-2)!)) * (5! / (2!(5-2)!))
= (4! / (2! * 2!)) * (5! / (2! * 3!))
= (4 * 3 * 2! / (2 * 2!)) * (5 * 4! / (2 * 3!))
= (4 * 3) * (5 / 2)
= 6 * 5
= 30
Therefore, there are 30 ways to choose the committee using combinations.
Approach 2: Using permutations
Another approach to solve this problem is by using permutations. We can calculate the number of permutations and then divide by the number of arrangements within the categories (students and faculty members).
Number of permutations for the committee = P(4, 4) = 4! / (4-4)!
= 4! / 0!
= 4! / 1
= 4!
However, within the committee, the students can arrange themselves in 2! ways and the faculty members can arrange themselves in 2! ways.
Therefore, the total number of ways to choose the committee using permutations = 4! / (2! * 2!) * 2! * 2!
= (4! / (2! * 2!)) * 2! * 2!
= (4 * 3 * 2! / (2 * 2!)) * 2 * 1 * 2
= (4 * 3) * 2
= 24
Therefore, there are 24 ways to choose the committee using permutations.
In conclusion, there are 30 ways to choose the committee using combinations and 24 ways to choose the committee using permutations.