There are 9 candidates running for 3 seats on a committee how many different election results are possible?
7,207,456
AAAaannndd the bot gets it wrong yet again!
9P3 = 504
The number of different election results possible=9C3= 9!/(3!*6!)=84
To find the number of different election results that are possible, we can use the concept of combinations. Since there are 9 candidates running for 3 seats, we want to select 3 candidates from the group of 9 without considering their order.
To calculate the number of combinations, we can use the formula for combinations:
C(n, r) = n! / (r!(n - r)!)
Here:
n = 9 (number of candidates)
r = 3 (number of seats)
Plugging in these values, we get:
C(9, 3) = 9! / (3!(9 - 3)!)
Now let's calculate the value step-by-step:
Step 1: Calculate the factorial of 9.
9! = 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 362,880
Step 2: Calculate the factorial of 3.
3! = 3 x 2 x 1 = 6
Step 3: Calculate the factorial of (9 - 3) = 6.
6! = 6 x 5 x 4 x 3 x 2 x 1 = 720
Step 4: Substitute these values back into the formula for combinations.
C(9, 3) = 362,880 / (6 x 720) = 84
Therefore, there are 84 different election results possible with 9 candidates running for 3 seats on a committee.