In order to find a viable solution for the planning committee made up of 12 students, with at least two students from each class (freshmen and sophomores), we can outline a few possible combinations.
Let's denote the number of freshmen as \( f \) and the number of sophomores as \( s \). According to the problem:
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The total number of students must equal 12: \[ f + s = 12 \]
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At least 2 students must be from each class: \[ f \geq 2 \] \[ s \geq 2 \]
Given these constraints, if we allocate 2 students from each class to start with (which meets the requirement of having at least 2 students from each class), we will have:
- 2 freshmen and 2 sophomores (total: 4 students)
- This leaves us with \( 12 - 4 = 8 \) students to fill.
Now, we can distribute these remaining 8 students among the two classes in any combination that keeps both \( f \) and \( s \) at a minimum of 2. Here are a few sample scenarios:
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Option A:
- \( f = 2 + 8 = 10 \) (2 freshmen + 8 more freshmen)
- \( s = 2 \)
- This results in \( f = 10 \) and \( s = 2 \): (10 Freshmen, 2 Sophomores)
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Option B:
- \( f = 4 \)
- \( s = 12 - 4 = 8 \)
- This results in \( f = 4 \) and \( s = 8 \): (4 Freshmen, 8 Sophomores)
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Option C:
- \( f = 5 \)
- \( s = 12 - 5 = 7 \)
- This results in \( f = 5 \) and \( s = 7 \): (5 Freshmen, 7 Sophomores)
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Option D:
- \( f = 6 \)
- \( s = 12 - 6 = 6 \)
- This results in \( f = 6 \) and \( s = 6 \): (6 Freshmen, 6 Sophomores)
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Option E:
- \( f = 7 \)
- \( s = 12 - 7 = 5 \)
- This results in \( f = 7 \) and \( s = 5 \): (7 Freshmen, 5 Sophomores)
Thus, any of these distribution scenarios, as long as they adhere to the rules set about minimum requirements and total must equal 12, represent viable solutions.
A final answer would then state that a solution could be any of those distributions, for instance, \( 6 \) freshmen and \( 6 \) sophomores would be a valid selection.
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