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Joined: 05 Jan 2006
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Answer
First, let's consider the different medal combinations that can be awarded to the 3 winners:
(1) If there are NO TIES then the three medals awarded are: GOLD, SILVER, BRONZE.
(2) What if there is a 2WAY tie?
If there is a 2WAY tie for FIRST, then the medals awarded are: GOLD, GOLD, SILVER.
If there is a 2WAY tie for SECOND, then the medals awarded are: GOLD, SILVER, SILVER.
There cannot be a 2WAY tie for THIRD (because exactly three medals are awarded in total).
(3) What if there is a 3WAY tie?
If there is a 3WAY tie for FIRST, then the medals awarded are: GOLD, GOLD, GOLD.
There are no other possible 3WAY ties.
Thus, there are 4 possible medal combinations:
(1) G, S, B (2) G, G, S (3) G, S, S (4) G, G, G
Now let's determine how many different ways each combination can be distributed. We'll do this by considering four runners: Albert, Bob, Cami, and Dora.
COMBINATION 1: Gold, Silver, Bronze
Gold Medal Silver Medal Bronze Medal
Any of the 4 runners can receive the gold medal. There are only 3 runners who can receive the silver medal. Why? One of the runners has already been awarded the Gold Medal. There are only 2 runners who can receive the bronze medal. Why? Two of the runners have already been awarded the Gold and Silver medals.
4 possibilities 3 possibilities 2 possibilities
Therefore, there are different victory circles that will contain 1 GOLD, 1 SILVER, and 1 BRONZE medalist.
COMBINATION 2: Gold, Gold, Silver.
Using the same reasoning as for Combination 1, we see that there are 24 different victory circles that will contain 2 GOLD medalists and 1 SILVER medalist. However, it is important to realize that these 24 victory circles must be reduced due to "overcounting."
To illustrate this, consider one of the 24 possible GoldGoldSilver victory circles:
Albert is awarded a GOLD. Bob is awarded a GOLD. Cami is awarded a SILVER.
Notice that this is the exact same victory circle as the following:
Bob is awarded a GOLD. Albert is awarded a GOLD. Cami is awarded a SILVER.
Each victory circle has been "overcounted" because we have been counting each different ordering of the two gold medals as a unique victory circle, when, in reality, the two different orderings consist of the exact same victory circle. Thus, the 24 victory circles must be cut in half; there are actually only 12 unique victory circles that will contain 2 GOLD medalists and 1 SILVER medalist. (Note that we did not have to worry about "overcounting" in Combination 1, because each of those 24 possibilities was unique.)
COMBINATION 3: Gold, Silver, Silver.
Using the same reasoning as for Combination 2, we see that there are 24 possible victory circles, but only 12 unique victory circles that contain 1 GOLD medalist and 2 SILVER medalists.
COMBINATION 4: Gold, Gold, Gold.
Here, once again, there are 24 possible victory circles. However, because all three winners are gold medalists, there has been a lot of "overcounting!" How much overcounting?
Let's consider one of the 24 possible GoldGoldGold victory circles:
Albert is awarded a GOLD. Bob is awarded a GOLD. Cami is awarded a GOLD.
Notice that this victory circle is exactly the same as the following victory circles:
AlbertGOLD, CamiGOLD, BobGOLD.
BobGOLD, AlbertGOLD, CamiGOLD.
BobGOLD, CamiGOLD, AlbertGOLD.
CamiGOLD, AlbertGOLD, BobGOLD.
CamiGOLD, BobGOLD, AlbertGOLD.
Each unique victory circle has actually been counted 6 times! Thus we must divide 24 by 6 to find the number of unique victory circles. There are actually only unique victory circles that contain 3 GOLD medalists.
FINALLY, then, we have the following:
(Combination 1) 24 unique GOLDSILVERBRONZE victory circles.
(Combination 2) 12 unique GOLDGOLDSILVER victory circles.
(Combination 3) 12 unique GOLDSILVERSILVER victory circles.
(Combination 4) 4 unique GOLDGOLDGOLD victory circles.
Thus, there are unique victory circles.
The correct answer is B.
