there are 9 candidates running for 3 seats on a committee. how many different election results are possible?
36
84
504
56***
You are "choosing" 3 from 9
The answer of 504 represents the number of permutations, that is, there are distinct places.
e.g. ABC is different from BAC
This is a combination question
number of ways to choose 3 from 9
= C(9,3)
= 84
C(9,3) = 9!/(3!6!)
Reiny is correct
I understand how you get to 504, but how do you get to 84? Can someone, please help me it would mean a lot!
Ohh ty. I got confused
what is the equation!!!!!!!!!!!! please tell mee !!!!!!!!!!!
No probs, it's pretty easy once you get the hang of it.
im a little late, but the equation reiny showed was 9 factorial divided by 3 factorial * 6 factorial.
To find the number of different election results possible, you can use the concept of permutations.
In this case, there are 9 candidates running for 3 seats on a committee, which means each seat can have a different candidate. To calculate the possible number of results, you need to find the number of ways to select 3 candidates out of 9 and arrange them on the seats.
The formula for calculating permutations is nPr = n! / (n-r)!, where n is the total number of items and r is the number of items you want to select.
In this case, n = 9 (number of candidates) and r = 3 (number of seats). So the calculation would be:
9P3 = 9! / (9-3)! = 9! / 6!
= (9 * 8 * 7 * 6!) / 6!
= (9 * 8 * 7)
= 504
Thus, there are 504 different election results possible for the 9 candidates running for 3 seats on the committee.
9!: 9x8x7
This is because they are just using 3 seats. Multiply that and you will get your answer. :)
Of course, you want a quick answer right? Well here is the answer, but I hope you know how to do this afterwards. . .
504