a judging panel will have 6 members chosen from 8 teachers and 10 students. there must be at least 3 students on the panel. in how many ways could there be
a) 3 students on the panel
b) 4 students on the panel
im not sure where to go on this one because from my understanding so far, the way to go about this would be (10C3) + (8C3) for A) at least
I believe you want 10C3 * 8C3, not +
B is similar
Why did the math book go to the therapist?
Because it had too many problems!
For this problem, let's consider the different possibilities:
a) If there are exactly 3 students on the panel, we can choose 3 students from the 10 available and 3 teachers from the remaining 8. So the number of ways would be (10C3) * (8C3).
b) If there are exactly 4 students on the panel, we can choose 4 students from the 10 available and 2 teachers from the remaining 8. So the number of ways would be (10C4) * (8C2).
Now, let me calculate that for you.
For part a) where there must be exactly 3 students on the panel, we can calculate it as follows:
The number of ways to choose 3 students from a group of 10 students is given by 10C3, which is calculated as:
10! / (3! * (10-3)!) = (10 * 9 * 8) / (3 * 2 * 1) = 120
Since there can be 3 students and the remaining 3 members can be chosen from the 8 teachers, the number of ways to choose 3 teachers from a group of 8 teachers is given by 8C3, which is calculated as:
8! / (3! * (8-3)!) = (8 * 7 * 6) / (3 * 2 * 1) = 56
To get the total number of ways for part a), we simply add these two possibilities together:
Total number of ways = 10C3 + 8C3 = 120 + 56 = 176
For part b) where there must be exactly 4 students on the panel, we can calculate it as follows:
The number of ways to choose 4 students from a group of 10 students is given by 10C4, which is calculated as:
10! / (4! * (10-4)!) = (10 * 9 * 8 * 7) / (4 * 3 * 2 * 1) = 210
Since there can be 4 students and the remaining 2 members can be chosen from the 8 teachers, the number of ways to choose 2 teachers from a group of 8 teachers is given by 8C2, which is calculated as:
8! / (2! * (8-2)!) = (8 * 7) / (2 * 1) = 28
To get the total number of ways for part b), we simply multiply these two possibilities together:
Total number of ways = 10C4 * 8C2 = 210 * 28 = 5,880
To find the number of ways to form the judging panel, we can use combination formulas.
a) To have exactly 3 students on the panel:
Step 1: Choose 3 students from 10. This can be calculated using the combination formula (nCr), where n is the number of students and r is the number of students selected for the panel. So, (10C3) = 10! / (3! * (10-3)!) = 120.
Step 2: Choose the remaining 3 panel members from the remaining 8 teachers and 7 students. This can be calculated as (15C3) = 15! / (3! * (15-3)!) = 455.
Step 3: Multiply the results from Step 1 and Step 2 to get the total number of ways: 120 * 455 = 54,600.
b) To have exactly 4 students on the panel:
Step 1: Choose 4 students from 10. This can be calculated as (10C4) = 10! / (4! * (10-4)!) = 210.
Step 2: Choose the remaining 2 panel members from the remaining 8 teachers and 6 students. This can be calculated as (14C2) = 14! / (2! * (14-2)!) = 91.
Step 3: Multiply the results from Step 1 and Step 2 to get the total number of ways: 210 * 91 = 19,110.
Therefore, there are 54,600 ways to form the judging panel with exactly 3 students and 19,110 ways to form the panel with exactly 4 students.