In a high jump competition with 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 is asking us to determine the number of ways that Mary, Dona, Sara, and Carla could be arranged in the top three spots, given that two of them tie for first place. This involves combinatorics and considering different scenarios of who ties and the order of the remaining individuals.
Answer
12
Answer for screen readers
12
Steps to Solve
- Identify possible pairs for the tie
There are 4 people, and we want to choose 2 of them to tie for first place. This can be done in $\binom{4}{2}$ ways.
- Calculate the number of ways to choose the pair
$\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3}{2 \times 1} = 6$
- Determine the number of possibilities for the third place
After choosing the pair that ties for first place, there are 2 people remaining. Either of these two people can take third place. Thus, there are 2 choices for third place.
- Calculate the total number of arrangements
Multiply the number of ways to choose the pair by the number of choices for third place.
Total arrangements = $6 \times 2 = 12$
12
More Information
The problem combines combinations (choosing the pair) with simple counting for the remaining position. It highlights how different counting principles can be combined to solve a problem.
Tips
A common mistake is to not realize that once the pair that ties is selected, the order of the remaining two people matters and only one can be in third place.
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