there are 9 candidates running for 3 seats....how many different election results are possible
I took the quiz its D- 84
45
To determine the number of different election results possible when 9 candidates are running for 3 seats, we can use the concept of combinations, specifically the "combinations with repetition" formula.
In this scenario, each of the 3 seats can be filled by any of the 9 candidates, and the same candidate can be chosen for multiple seats. This means we need to calculate the number of different combinations of candidates for the 3 seats.
The formula for combinations with repetition is given by:
C(n + r - 1, r)
where n is the number of candidates (9) and r is the number of seats (3).
Substituting the values into the formula:
C(9 + 3 - 1, 3) = C(11, 3)
Now, we need to calculate the value of C(11, 3).
Using the combination formula:
C(11, 3) = 11! / (3! * (11-3)!)
Simplifying:
C(11, 3) = 11! / (3! * 8!)
Calculating the factorials:
11! = 11 * 10 * 9 * 8!
3! = 3 * 2 * 1
Substituting the values:
C(11, 3) = (11 * 10 * 9 * 8!) / (3 * 2 * 1 * 8!)
Canceling out common terms:
C(11, 3) = (11 * 10 * 9) / (3 * 2 * 1)
Calculating:
C(11, 3) = 165
Therefore, there are 165 different election results possible when 9 candidates are running for 3 seats.