In a high jump competition among Mary, Dona, Sara, and Carla, if two of them tie for first place, how many ways could they be arranged in the top three spots?
Understand the Problem
The question involves combinatorics and permutations. We need to figure out the number of ways two people can tie for first place out of a group of four, and then consider the possible arrangements for the top three positions. This requires calculating combinations and permutations to arrive at the final answer.
Answer
12
Answer for screen readers
12
Steps to Solve
- Choose the two winners
We need to choose 2 people out of 4 to tie for first place. This is a combination problem, since the order in which we choose the two people doesn't matter. The number of ways to choose 2 people out of 4 is given by the combination formula:
$$ {4 \choose 2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3 \times 2 \times 1}{(2 \times 1)(2 \times 1)} = \frac{24}{4} = 6 $$
- Determine the possible arrangements for the top 3
Once we have the two people who tie for first place, we need to determine who comes in third. Since two people tied for first, there are two 'first place' positions filled. There are now 2 people remaining who could come in third. So there are 2 choices for the third position.
- Calculate the total number of ways
Multiply the number of ways to choose the two winners by the number of possible choices for third place to determine the total number of possible outcomes.
$$ \text{Total ways} = {4 \choose 2} \times 2 = 6 \times 2 = 12 $$
12
More Information
The problem combines combinations (selecting the two winners) and basic counting principles to arrive at the final answer.
Tips
A common mistake is to confuse combinations and permutations. In this problem, the order in which the two people are chosen to tie for first place doesn't matter, so it's a combination, not a permutation. Another mistake is forgetting that after two people tie for first place, there are only two remaining candidates for third place.
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