1.Find the number of pairwise comparisons that would take place in an election with 13 candidates?
2.If there are 171 total pairwise comparisons in a different election, how many candidates are running?
each of the 13 candidates is paired with each of the other 12
... BUT ... A paired with B is the same as B paired with A
1. 13 * 12 / 2 = ?
2. c = (171 * 2) / (c - 1) ... c^2 - c - 342 = 0
So 342 people are running for the second question or would it be 0?
c is the number of candidates
you need to solve the quadratic to find the answer
factoring ... (c + 18) (c - 19) = 0
1. To find the number of pairwise comparisons that would take place in an election with 13 candidates, we can use the concept of combinations. In this case, we want to select 2 candidates at a time from a group of 13 candidates.
The formula to calculate the number of combinations is given by:
nCk = n! / (k! * (n - k)!)
where n is the total number of candidates and k is the number of candidates selected at a time.
Using this formula, we plug in the values:
n = 13 and k = 2.
13C2 = 13! / (2! * (13-2)!)
= 13! / (2! * 11!)
= (13 * 12) / (2 * 1)
= 78
Therefore, in an election with 13 candidates, there would be a total of 78 pairwise comparisons.
2. If there are 171 total pairwise comparisons in a different election, we can reverse the calculation process to find the number of candidates running.
Again, using the formula for combinations:
nCk = n! / (k! * (n - k)!)
This time, we have n unknown and k = 2.
Also, we are given nCk = 171.
171 = n! / (2! * (n - 2)!)
By rearranging the formula, we get:
n! = 171 * (2! * (n - 2)!)
Now, let's try different values for n to find a solution.
For n = 10:
10! = 171 * (2! * (10 - 2)!)
10! = 171 * (2 * 8!)
10! = 171 * 2 * 8!
10! = 342 * 8!
10! = 342 * 40320
10! = 13789440
As we can see, 10! is not equal to 13789440, which means n = 10 is not a valid solution.
Let's try n = 11 instead:
11! = 171 * (2! * (11 - 2)!)
11! = 171 * (2 * 9!)
11! = 171 * 2 * 9!
11! = 342 * 9!
11! = 342 * 362880
11! = 1245404160
Again, 11! is not equal to 1245404160, so n = 11 is not a valid solution.
Now, let's try n = 12:
12! = 171 * (2! * (12 - 2)!)
12! = 171 * (2 * 10!)
12! = 171 * 2 * 10!
12! = 342 * 10!
12! = 342 * 3628800
12! = 12454041600
Finally, 12! is equal to 12454041600, which means n = 12 is a valid solution.
Therefore, in the election with 171 total pairwise comparisons, there are 12 candidates running.