The Medal Count
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
And here is your answer: (Please find my comments at the end)
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, Bronze
Gold 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 x 3 x 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 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.
-----> My Observation:
When a circle is formed with 3 different people, there are 2 ways of forming the circle 1,2,3 and 1,3,2 and it is a permutation calculation (n-1)!. So the answer should be 104.
That's my understanding of the problem and slution. Please correct me if I am wrong.
Also, another way of doing this problem is :
We can select 3 people out of 4 in 4 ways.
There are 3*2*1 = 6 different ways of giving G,S,B different medals to 3 people.
There are 3 different ways of giving G,G,S different medal to 3 people.
There are 3 different ways of giving G,B,B different medal to 3 people.
There is 1 way of giving G,G,G to 3 people.
So total 6+3+3+1 = 13 ways of giving medals.
There are 2 different ways in which 3 people can form circles.
So total number of ways is 4*13*2 = 104.
GMAT Forum
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- StaceyKoprince
- ManhattanGMAT Staff
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- christiancryan
- Course Students
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- Joined: Thu Jul 31, 2003 10:44 am
Hi Anadi,
I apologize -- you had contacted us directly some time ago about this problem, we wrote up a response but in our internal handoff apparently we dropped the ball. Very sorry about that.
Here is the response:
1) The "victory circle" is not a literal circle. The meaning of "circle" here is "group." So your reference to (n-1)! as the formula to calculate the permutations of positions in a circle, while correct in its own context, is not applicable here.
2) In your second solution, you didn't take into account the 4th person in doing the permutation calculations.
3) The proper way to do this problem is outlined in our solution:
a) There are 4! = 24 ways to distribute 3 distinct medals (G, S, B) to 4 people. The anagram method works here: use the sign N for the person who does NOT receive a medal. So then this part of the question becomes, how many ways can you rearrange the letters GSBN to form different words? The answer is 4! = 24 ways, since all the symbols are different.
b) To handle the case GGB (2 golds and a bronze), you rearrange GGBN. The answer is 4!/2!. The 2! takes into account the repeats created by having 2 identical medals; dividing by 2! appropriately reduces the answer (for this case) to 12 ways.
c) Likewise, there are 12 ways to rearrange GSSN (1 gold and 2 silvers).
d) Finally, there are 4 ways to rearrange GGGN (3 golds and that’s it).
I hope this makes sense. Please reply if you have further questions.
Best,
Chris
I apologize -- you had contacted us directly some time ago about this problem, we wrote up a response but in our internal handoff apparently we dropped the ball. Very sorry about that.
Here is the response:
1) The "victory circle" is not a literal circle. The meaning of "circle" here is "group." So your reference to (n-1)! as the formula to calculate the permutations of positions in a circle, while correct in its own context, is not applicable here.
2) In your second solution, you didn't take into account the 4th person in doing the permutation calculations.
3) The proper way to do this problem is outlined in our solution:
a) There are 4! = 24 ways to distribute 3 distinct medals (G, S, B) to 4 people. The anagram method works here: use the sign N for the person who does NOT receive a medal. So then this part of the question becomes, how many ways can you rearrange the letters GSBN to form different words? The answer is 4! = 24 ways, since all the symbols are different.
b) To handle the case GGB (2 golds and a bronze), you rearrange GGBN. The answer is 4!/2!. The 2! takes into account the repeats created by having 2 identical medals; dividing by 2! appropriately reduces the answer (for this case) to 12 ways.
c) Likewise, there are 12 ways to rearrange GSSN (1 gold and 2 silvers).
d) Finally, there are 4 ways to rearrange GGGN (3 golds and that’s it).
I hope this makes sense. Please reply if you have further questions.
Best,
Chris
3 posts Page 1 of 1