Ms. Bell's mathematics class consists of 15 sophomores, 9 juniors, and 6 seniors. How many different ways can Ms. Bell create a 3-member committee of sophomores if each sophomore has an equal chance of being selected?
Question
Ms. Bell's mathematics class consists of 15 sophomores, 9 juniors, and 6 seniors. How many different ways can Ms. Bell create a 3-member committee of sophomores if each sophomore has an equal chance of being selected?
Solution
To solve this problem, we need to use the concept of combinations in mathematics. A combination is a selection of items without considering the order.
The formula for combinations is:
C(n, r) = n! / [r!(n-r)!]
where:
- n is the total number of items,
- r is the number of items to choose,
- "!" denotes a factorial, meaning the product of all positive integers up to that number.
In this case, we want to find the number of ways to choose a 3-member committee from 15 sophomores. So, n = 15 (the total number of sophomores) and r = 3 (the number of sophomores to choose for the committee).
Substituting these values into the formula gives:
C(15, 3) = 15! / [3!(15-3)!]
Calculating the factorials:
15! = 15 × 14 × 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 3! = 3 × 2 × 1 (15-3)! = 12! = 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
Substituting these values back into the formula gives:
C(15, 3) = (15 × 14 × 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1) / [(3 × 2 × 1) × (12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)]
Simplifying this expression, we find that C(15, 3) = 455.
So, there are 455 different ways that Ms. Bell can create a 3-member committee of sophomores.
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