In a high jump competition with Mary, Dona, Sara, and Carla, if two of the four tie for first place, how many ways could they be arranged in the top three spots?

Steps to Solve

1. Determine the number of ways to choose the pair that ties for first place

There are 4 athletes, and we need to choose 2 of them to tie for first. This is a combination problem, so we use the combination formula:

$$C(n, k) = \frac{n!}{k!(n-k)!}$$

In this case, $n = 4$ and $k = 2$, so:

$$C(4, 2) = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3 \times 2 \times 1}{(2 \times 1)(2 \times 1)} = \frac{24}{4} = 6$$

So, there are 6 possible pairs of athletes who can tie for first place.

2. Determine the number of athletes that can take the third spot

Since two athletes tied for first, there are $4 - 2 = 2$ athletes remaining who could potentially take third place.

3. Calculate the total number of possible outcomes

For each of the 6 possible pairs that tie for first, there are 2 possible athletes who can take third place. Therefore, the total number of ways the competition can end is:

$$6 \times 2 = 12$$