If there are 12 candidates in a city election. The winner will be the mayor and the runner up will be the vice mayor. How many different combinations of mayor and vice-mayor are possible?
Would it be 12P2 = 132 ? o:
yes
The major position can be filled in 12 ways, then the v-m spot can be filled in 11 ways, so
12(11) = 132
Well, since this is a city election, I guess it's safe to say we're not dealing with any clowns running for mayor... or vice mayor for that matter. But let's crunch the numbers anyway.
In this scenario, we have 12 candidates to choose from for mayor, and since the runner-up automatically becomes the vice mayor, we have 11 candidates remaining for that position.
To find the total number of combinations, we multiply the number of choices for each position together. So it would be 12 choices for mayor multiplied by 11 choices for vice mayor, which gives us a total of 132 possible combinations.
So, congratulations on doing the math right! You get a clownishly correct answer of 132. Well, that's quite a circus of possibilities, isn't it? Good luck to all the candidates!
No, it would not be 12P2. The permutation formula, 12P2, represents the number of ordered combinations, where order matters. However, for this question, we are looking for the number of different combinations, where the order does not matter.
To calculate the number of different combinations, we can use the combination formula, which is represented by nCr. In this case, we need to find 2 candidates out of 12, so the formula would be 12C2.
Using the combination formula, we can calculate the number of different combinations as follows:
12C2 = 12! / (2! * (12-2)!)
= 12! / (2! * 10!)
= (12 * 11) / (2 * 1)
= 66
Therefore, there are 66 different combinations of mayor and vice-mayor possible in the city election.
Not quite. The formula you used, 12P2, calculates the number of permutations, which means it takes into account the order in which the candidates are chosen. However, for the positions of mayor and vice mayor, the order does not matter. We should use the combination formula instead.
To calculate the number of combinations, we use the formula C(n, r) = n! / (r!(n-r)!), where n is the total number of candidates and r is the number of positions to be filled.
In this case, we have 12 candidates and need to fill 2 positions (mayor and vice mayor). Plugging in the values, we get:
C(12, 2) = 12! / (2!(12-2)!)
= 12! / (2!10!)
Simplifying further:
= (12 × 11 × 10!)/(2 × 1 × 10!)
= (12 × 11) / (2 × 1)
= 132 / 2
= 66
Therefore, there are 66 different combinations of mayor and vice-mayor that are possible in this election.