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# In a 4 person race, medals are awarded to the fastest 3

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In a 4 person race, medals are awarded to the fastest 3 [#permalink]

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11 Apr 2006, 22:13
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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
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11 Apr 2006, 22:15
This is One old MGMT Challage!

OA is B and I have no idea how to get this! I tried what ever I can think off! That includes reading explaination! but still I don't get it!
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12 Apr 2006, 12:49
chiragr wrote:
This is One old MGMT Challage!

OA is B and I have no idea how to get this! I tried what ever I can think off! That includes reading explaination! but still I don't get it!
\

Can you put explanation here
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12 Apr 2006, 12:57
hmmmm....
I'm going to pick A.
Not very mathematical, but-

Out of three winners - there are 6 possible permutations (think trifecta @ the track )

and there are 4 ways (3C4) to choose those three winners.

6*4 = 24

So I guess the "mathematical" way is 3P3 * 3C4 = 3! * 4!/(3!*1!) = 24
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12 Apr 2006, 13:04
nevermind - i read too fast - didn't count the "ties" in there...
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12 Apr 2006, 15:03
Heres how I tried.

There are 4 possible ways in which the 3 positions for awards can be done

F S T
F F S
F S S
F F F

(F- First position S Second position and T third position)

we have 4 players so we have to select 3 from 4 so 4C3 ways = 4 --(i)

Now come to the arrangements

F S T 3! = 6 ways
F F S and F S S in 3!/2! ways for each one which is 3+3 = 6 ways
F F F in 1 way 3!/3! = 1

So 6+6+1 = 13

Multiplied by (i) which is the selections = 13*4 =52 ways....
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12 Apr 2006, 15:24
gmat_crack wrote:
chiragr wrote:
This is One old MGMT Challage!

OA is B and I have no idea how to get this! I tried what ever I can think off! That includes reading explaination! but still I don't get it!
\

Can you put explanation here

I am waiting for few more to reply..
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12 Apr 2006, 23:31
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 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 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 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 unique victory circles.

12 Apr 2006, 23:31
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