If two of the four high jump athletes Mary, Dona, Sara, and Carla tie for first place, how many ways could they be arranged in the top three spots?
Understand the Problem
The question asks to find the number of arrangements of the top 3 athletes in a high jump competition given that two of the four athletes tie for first place. This requires considering the different pairs that can tie and then the possible arrangements of the remaining athletes.
Answer
12
Answer for screen readers
12
Steps to Solve
- Determine the number of ways to choose the pair of athletes that tie for first place.
Since there are 4 athletes, we need to choose 2 of them to tie for first place. This can be done in $\binom{4}{2}$ ways.
$$ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3}{2 \times 1} = 6 $$
- Determine the number of ways to arrange the remaining two athletes.
After selecting the two athletes who tie for first place, there are 2 athletes remaining. One of these athletes will be in third place. There are $2$ choices for the third place athlete. Once the third-place athlete is selected, there is only $1$ athlete remaining, and they will be in fourth place. Therefore, there are $2 \times 1 = 2! = 2$ ways to arrange the remaining two athletes in third and fourth place.
- Calculate the total number of possible arrangements.
Multiply the number of ways to choose the pair of athletes that tie for first place by the number of ways to arrange the remaining athletes for third and fourth place.
$$ 6 \times 2 = 12 $$
12
More Information
There are 12 possible arrangements of the top 3 athletes, given that two of the four athletes tie for first place. The other two athletes are distinguishable.
Tips
A common mistake would be not recognizing to use combinations to find the different pairs of athletes, which means the order in which the pair of athletes are selected does not matter (as both are tied for first place).
AI-generated content may contain errors. Please verify critical information
