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Step 1: Understand the Problem
We have three classes with the following number of students:
Class 10: 5 students
Class 11: 6 students
Class 12: 8 students
We want to select 10 students in total, satisfying these conditions:
At least 2 students from each class.
At most 5 students from Class 10 and Class 11 combined (i.e., from the total 11 students of classes 10 and 11, we select at most 5).
Step 2: Enumerate Possible Distributions
Let the number of selected students from Class 10, Class 11, and Class 12 be denoted by (x, y, z) respectively. We need:
x + y + z = 10
x ≥ 2, y ≥ 2, z ≥ 2
x + y ≤ 5
Checking these constraints, the valid distributions (x, y, z) are:
(2, 2, 6)
(2, 3, 5)
(3, 2, 5)
Step 3: Calculate Number of Ways for Each Distribution
To select x students out of 5 in Class 10, we use the combination ^{5}C_{x} , for y students out of 6 in Class 11, we use ^{6}C_{y} , and for z students out of 8 in Class 12, we use ^{8}C_{z} . We then multiply these combinations because the selections are independent.
Distribution (2, 2, 6):
^{5}C_{2} \times ^{6}C_{2} \times ^{8}C_{6}
Distribution (2, 3, 5):
^{5}C_{2} \times ^{6}C_{3} \times ^{8}C_{5}
Distribution (3, 2, 5):
^{5}C_{3} \times ^{6}C_{2} \times ^{8}C_{5}
Step 4: Compute Each Term
Compute each combination value:
^{5}C_{2} = 10
^{6}C_{2} = 15
^{6}C_{3} = 20
^{5}C_{3} = 10
^{8}C_{5} = 56
^{8}C_{6} = 28
Hence:
For (2, 2, 6):
^{5}C_{2} \times ^{6}C_{2} \times ^{8}C_{6} = 10 \times 15 \times 28 = 4200
For (2, 3, 5):
^{5}C_{2} \times ^{6}C_{3} \times ^{8}C_{5} = 10 \times 20 \times 56 = 11200
For (3, 2, 5):
^{5}C_{3} \times ^{6}C_{2} \times ^{8}C_{5} = 10 \times 15 \times 56 = 8400
Step 5: Sum All Possible Ways
The total number of ways to choose the students is the sum of all three distributions:
4200 + 11200 + 8400 = 23800
Step 6: Find the Value of k
According to the problem, the number of ways is given to be 100k . So we set:
100k = 23800
Solving for k gives:
k = 238
Final Answer
k = 238
