ziyuenlau wrote:In a 4 person race, medals are awarded to the fastest 3 runners. The first-place runner receives a gold medal, the second-place runner receives a silver medal, and the third-place runner receives a bronze medal. In the event of a tie, the tied runners receive the same color medal. (For example, if there is a two-way tie for first-place, the top two runners receive gold medals, the next-fastest runner receives a silver medal, and no bronze medal is awarded). Assuming that exactly three medals are awarded, and that the three medal winners stand together with their medals to form a victory circle, how many different victory circles are possible?

A) 24
B) 52
C) 96
D) 144
E) 648

The OA implies the following that a victory circle is simply a victory GROUP:
the three fastest runners, along with their respective medals.

Let G = gold, S = silver, and B = bronze.

Since ties are allowed, the following combinations of medal winners are possible:
Case 1: G-G-G (all three runners tie one another)
Case 2: G-G-S (the two fastest runners tie each other)
Case 3: G-S-S (a two-way tie for second place)
Case 4: G-S-B (no ties).

No other cases are possible.
Since S will be awarded only after G is awarded, S-S-S, S-S-B and S-B-B are not viable.
Since B will be awarded only if G and S are both awarded, G-B-B and B-B-B are not viable.

Case 1: G-G-G
From the 4 runners, the number of ways to choose a group of 3 to receive gold medals = 4C3 = (4*3*2)/(3*2*1) = 4.

Case 2: G-G-S
From the 4 runners, the number of ways to choose a pair to receive gold medals = 4C2 = (4*3)/(2*1) = 6.
Number of options for S = 2. (Either of the two remaining runners.)
To combine these options, we multiply:
6*2 = 12.

Case 3: G-S-S
From the 4 runners, the number of ways to choose a pair to receive silver medals = 4C2 = (4*3)/(2*1) = 6.
Number of options for G = 2. (Either of the two remaining runners.)
To combine these options, we multiply:
6*2 = 12.

Case 4: G-S-B
Number of options for G = 4. (Any of the runners.)
Number of options for S = 3. (Any of the 3 remaining runners.)
Number of options for B = 2. (Either of the 2 remaining runners.)
To combine these options, we multiply:
4*3*2 = 24.

Total ways = Case 1 + Case 2 + Case 3 + Case 4 = 4+12+12+24 = 52.

The correct answer is B.