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# Victory circles- Combination/counting

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Victory circles- Combination/counting [#permalink]  20 Nov 2009, 02:53
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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?
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Re: Victory circles- Combination/counting [#permalink]  20 Nov 2009, 04:26
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?

Hi,

This is my method for solving this problem.

The possible ways for exactly 3 medals to be awarded are :

CASE 1 : Gold, Silver, Bronze

4 choices for gold, 3 choices for silver (because one already has gold), and 2 choices for bronze = 4*3*2 = 24

CASE 2 : Gold, Gold, Silver

From 4 we have to first select two people for gold and then from 2 we have to select 1 person for silver : (4C2)*(2C1) = 12

Note: We use combinations to avoid repetitions. It is the same as using the formula for the number of permutations for n different things such that q of one type are similar, r of another type are similar, etc. (\frac{n!}{q!r!...)}

CASE 3 : Gold, Silver, Silver

From 4 we have to first select 1 person for gold and then from 3 we have to select 2 people for silver : (4C1)*(3C2) = 12

CASE 4 : Gold, Gold, Gold

From 4 we have to select 3 people for the gold medal : (4C3) = 4

Note : We cannot have a case where more than 1 bronze medal is awarded as there would already be one gold and silver awarded before that and we cannot have more than 3 medalists.

Therefore, the total number of ways of forming groups of three medalists is 24 + 12 + 12 + 4 = 52

Now, I am not sure whether victory circle refers to them actually standing in a circle or not (its usually not taken as a literal meaning so my first guess would be no). However, if 52 does not match any of the answer choices then we would have to assume that they are actually standing in a circle and therefore multiply this by (n-1)!

Therefore if they were actually standing in a circle, the answer would be 52*(3-1)! = 104
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Re: Victory circles- Combination/counting [#permalink]  20 Nov 2009, 04:52
Srihari,

OA is indeed 52.
thanks for the soln.
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Re: Victory circles- Combination/counting [#permalink]  13 May 2011, 03:31
4*3*2 + 4c2* 2c1 + 4c1* 3c2 + 4c3

52 is a clean shot.
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Re: Victory circles- Combination/counting [#permalink]  13 May 2011, 11:44
There will always be a Gold Winnner
So Possible combination are

GSB= 4*3*2 =24
GGS= (4C2)*(2C1)= 6*2=12
GSS= (4C1)*(3C2)= 4*3=12
GGG= (4C3)=4

Total = 24 + 12 + 12 + 4 = 52
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Re: Victory circles- Combination/counting   [#permalink] 13 May 2011, 11:44
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