kirankp 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
OFFICIAL SOLUTION
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 2-WAY tie?
--If there is a 2-WAY tie for FIRST, then the medals awarded are: GOLD, GOLD, SILVER.
--If there is a 2-WAY tie for SECOND, then the medals awarded are: GOLD, SILVER, SILVER.
--There cannot be a 2-WAY tie for THIRD (because exactly three medals are awarded in total).
(3) What if there is a 3-WAY tie?
--If there is a 3-WAY tie for FIRST, then the medals awarded are: GOLD, GOLD, GOLD.
--There are no other possible 3-WAY 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, BronzeGold Medal Any of the 4 runners can receive the gold medal. (
4 possibilities)
Silver Medal There are only 3 runners who can receive the silver medal. Why? One of the runners has already been awarded the Gold Medal. (
3 possibilities)
Bronze 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. (
2 possibilities)
Therefore, there are \(4*3*2=24\) 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 Gold-Gold-Silver
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 Gold-Gold-Gold
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:
Albert-GOLD, Cami-GOLD, Bob-GOLD.
Bob-GOLD, Albert-GOLD, Cami-GOLD.
Bob-GOLD, Cami-GOLD, Albert-GOLD.
Cami-GOLD, Albert-GOLD, Bob-GOLD.
Cami-GOLD, Bob-GOLD, Albert-GOLD.
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 \(\frac{24}{6}=4\) unique
victory circles that contain 3 GOLD medalists.
FINALLY, then, we have the following:
(Combination 1) 24 unique GOLD-SILVER-BRONZE
victory circles.
(Combination 2) 12 unique GOLD-GOLD-SILVER
victory circles.
(Combination 3) 12 unique GOLD-SILVER-SILVER
victory circles.
(Combination 4) 4 unique GOLD-GOLD-GOLD
victory circles.
Thus, there are \(24+12+12+4=52\) unique
victory circles.
The correct answer is
B.
Bunuel, Original source is
Manhattan Prep. Please help to edit. Thank you.
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