A committee must be formed with 2 teachers and 4 students. If there are 9 teachers to choose from, and 9 students, how many different ways could the committee be made?
Well, if it's anything like forming a committee of clowns, it's bound to be quite the circus!
To figure out the number of different ways the committee can be formed, we can use combinations. We need to choose 2 teachers out of 9 available options, and 4 students out of 9 options as well.
For the teachers, we can calculate the number of combinations by using the formula:
C(n, r) = n! / (r!(n-r)!)
where n is the total number of options and r is the number we choose.
So, for the teachers, we have:
C(9, 2) = 9! / (2!(9-2)!) = 9! / (2!7!) = (9 * 8) / (2 * 1) = 36
Now, we do the same for the students:
C(9, 4) = 9! / (4!(9-4)!) = 9! / (4!5!) = (9 * 8 * 7 * 6) / (4 * 3 * 2 * 1) = 126
To find the total number of different ways the committee can be formed, we multiply these two results:
36 * 126 = 4,536
So, there are 4,536 different ways the committee can be formed. And with that many clowns involved, you can bet it'll be a laugh riot!
To solve this problem, we can use the combination formula, which is represented as:
C(n, k) = n! / (k! * (n - k)!)
Where:
- n is the total number of options for a category,
- k is the number of options that will be selected, and
- "!" represents the factorial of a number.
In this case, we have 9 teachers to choose from for the two teacher positions, which gives us C(9, 2) options. We also have 9 students to choose from for the four student positions, which gives us C(9, 4) options.
Using the combination formula, we can calculate the number of ways the committee can be formed:
C(9, 2) = 9! / (2! * (9-2)!)
= 9! / (2! * 7!)
= (9 * 8 * 7!) / (2!)
= (9 * 8) / (2 * 1)
= 72 / 2
= 36
C(9, 4) = 9! / (4! * (9-4)!)
= 9! / (4! * 5!)
= (9 * 8 * 7 * 6 * 5!) / (4! * 5!)
= (9 * 8 * 7 * 6) / (4 * 3 * 2 * 1)
= 3024 / 24
= 126
Thus, the number of different ways the committee can be made is the product of C(9, 2) and C(9, 4):
Number of ways = C(9, 2) * C(9, 4)
= 36 * 126
= 4536
Therefore, there are 4,536 different ways the committee can be made.
To find the number of different ways the committee can be formed, we can use the concept of combinations.
First, we need to select 2 teachers from a group of 9 teachers. We can calculate this using the formula for combinations: nCr = n! / (r! * (n - r)!), where n is the total number of objects, and r is the number of objects to be chosen.
In this case, we have 9 teachers and we need to choose 2 of them, so the calculation becomes: 9C2 = 9! / (2! * (9 - 2)!) = (9 * 8) / (2 * 1) = 9 * 4 = 36.
Next, we need to select 4 students from a group of 9 students. Again, we can calculate this using the formula for combinations: 9C4 = 9! / (4! * (9 - 4)!) = (9 * 8 * 7 * 6) / (4 * 3 * 2 * 1) = 9 * 7 = 63.
Finally, we can find the total number of different ways the committee can be formed by multiplying the two results together: 36 * 63 = 2,268.
Therefore, there are 2,268 different ways the committee can be made.