Answer: 81
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Step-by-step explanation:
2 freshmen + 3 sophomores + 2 juniors + 2 seniors = 9 students total.
Let's consider the cases where we have the same number of juniors and seniors. We'll then take the complement of this to get the final answer.
- Case A will look at having 1 of each junior and senior.
- Case B will look at having 2 each of juniors and seniors.
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Case A: There is 1 junior and 1 senior
There are 2 ways to pick a junior and 2 ways to pick a senior. That's 2*2 = 4 ways so far.
Then we have 9-4 = 5 students left to pick from (i.e. 2 freshmen+3 sophomores = 5 students left) and we have 5-2 = 3 seats to fill. Order doesn't matter on the committee since each person has equal rank.
We use the nCr combination formula.
n = 5 students
r = 3 seats to fill
n C r = (n!)/(r!(n-r)!)
5 C 3 = (5!)/(3!*(5-3)!)
5 C 3 = (5!)/(3!*2!)
5 C 3 = (5*4*3!)/(3!*2!)
5 C 3 = (5*4)/(2!)
5 C 3 = (5*4)/(2*1)
5 C 3 = (20)/(2)
5 C 3 = 10
There are 10 ways to fill the remaining 3 seats when we pick from the freshmen and sophomores only.
To recap everything so far:
- 4 ways to pick the 1 junior and 1 senior
- 10 ways to pick the 3 other students (freshmen + sophomores)
Therefore, we have 4*10 = 40 different combinations possible for case A. We'll refer to this value later.
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Case B: We pick 2 juniors and 2 seniors
Since there 2 juniors to pick from, and 2 junior seats to fill, there's only 1 way to do this. Likewise, there's only 1 way to pick the 2 seniors to fill the 2 seats.
In total so far there is 1*1 = 1 way to pick the 2 juniors and 2 seniors in any order you like.
Then we have 9-4 = 5 students left that are freshmen or sophomores. This is the number of choices we have for the final 5th seat.
We have 1*5 = 5 ways to have case B happen.
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Summary so far:
- 40 ways to do case A
- 5 ways to do case B
- 40+5 = 45 ways to do either case.
There are 45 ways to have the same number of juniors as seniors (either 1 of each or 2 of each).
Now we must calculate the number of total combinations possible on this committee. We'll turn to the nCr formula again.
n = 9 students
r = 5 seats
n C r = (n!)/(r!(n-r)!)
9 C 5 = (9!)/(5!*(9-5)!)
9 C 5 = (9!)/(5!*4!)
9 C 5 = (9*8*7*6*5!)/(5!*4!)
9 C 5 = (9*8*7*6)/(4!)
9 C 5 = (9*8*7*6)/(4*3*2*1)
9 C 5 = (3024)/(24)
9 C 5 = 126
There are 126 different five-person committees possible.
Of those 126 committees, 45 of them consist of cases where we have the same number of juniors and seniors.
That must mean there are 126-45 = 81 combinations where we do not have the same number of juniors and seniors. This is where the concept of "complement" comes in.