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

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03 Dec 2009, 07:52

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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?

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

Possible scenarios are:

1. Gold/Silver/Bronze/No medal (no ties) - 4!=24; 2. Gold/Gold/Silver/No medal - 4!/2!=12; 3. Gold/Silver/Silver/No medal - 4!/2!=12; 4. Gold/Gold/Gold/No medal - 4!/3!=4.

1. Gold/Silver/Bronze/No medal (no ties) - 4!=24; 2. Gold/Gold/Silver/No medal - 4!/2!=12; 3. Gold/Silver/Silver/No medal - 4!/2!=12; 4. Gold/Gold/Gold/No medal - 4!/3!=4.

Total: 24+12+12+4=52

Answer: B.

Sorry, but i dont see how this answers the "how many different victory circles are possible" part of the question?

If we are concerned only with medals in the victory circle:

Possible medal combinations are: GSB, GSS, GGS, GGG. Possible victory circles [N.B. removing rotationally symmetric permutations]:

GSB: = 2 GSS: = 1 GGS: = 1 GGG: = 1

So, total possible victory circles are 5.

If we are concerned with combinations of people and medals in the victory circle:

Possible permutation of winners (top 3) = 4P3 = 24 Possible victory circles:

1. Gold/Silver/Bronze/No medal (no ties) - 4!=24; 2. Gold/Gold/Silver/No medal - 4!/2!=12; 3. Gold/Silver/Silver/No medal - 4!/2!=12; 4. Gold/Gold/Gold/No medal - 4!/3!=4.

Total: 24+12+12+4=52

Answer: B.

Sorry, but i dont see how this answers the "how many different victory circles are possible" part of the question?

If we are concerned only with medals in the victory circle:

Possible medal combinations are: GSB, GSS, GGS, GGG. Possible victory circles [N.B. removing rotationally symmetric permutations]:

GSB: = 2 GSS: = 1 GGS: = 1 GGG: = 1

So, total possible victory circles are 5.

If we are concerned with combinations of people and medals in the victory circle:

Possible permutation of winners (top 3) = 4P3 = 24 Possible victory circles:

Please can i get some advice on where i went wrong with both of these approaches?

I don't quite understand your solution. Here is the logic behind mine:

We have four possible patterns (GSBN, GSSN, GGSN, GGGN) and four persons (a, b, c, d). Each victory circle is made by assigning pattern letters to these persons as follows: Firs pattern: GSBN a--b--c--d G--S--B--N G--B--S--N G--B--N--S ... So how many victory circles are possible for the first pattern? It would be the # of permutations of the letters GSBN, as these are four distinct letters, the # is 4!=24.

The same for other patterns: Second pattern: GSSN. # of permutations is 4!/2!=12 (as there are four letters out of which 2 are the same).

Third pattern: GGSN. # of permutations is 4!/2!=12.

Fourth pattern: GGGN. # of permutations is 4!/3!=4. a--b--c--d G--G--G--N G--G--N--G G--N--G--G N--G--G--G Here you can see that these four victory circles are all different and the same will be for other patterns.

24+12+12+4=52

Hope it's clear.

RaviChandra wrote:

what if 2 people come first(i.e. Gold) and another 2 come second position(silver)

We are told in the stem that only 3 medals will be awarded, so we should take this as a fact. _________________

Thanks for the clarification - I'm still not comfortable with the answer and i have three more questions:

1) Is a "victory circle" an actual circle and therefore are there rotational symmetry considerations that need to be taken into account when calculating arrangements. [I've never heard of this term before - so apologies if this is a silly question?]

2) The question asks: "three medal winners stand together with their medals to form a victory circle" so are we calculating permutations with to many people in the 'victory circle'? [This is the only flaw i'm pretty sure about which takes the number of permutation down to 13 - which is not an answer]

3)The questions asks: "the three medal winners stand together with their medals" so should we be calculating permutations of people and medals and not discounting duplicate medals? [Potentially tenuous but needed to get to an answer - see below]

Assuming that (1) a 'victory circle' is not a circle but a straight line, (2) that we are only concerned with medal holders, and (3) person A with a silver medal can be considered different to person B with a silver medal (in a scenario where they both have one) then:

Possible medal combinations: GSB, GSS, GGS, GGG

GSB victory circle permutations = 4P3 = 24 GSS victory circle permutations = 4P3 = 24 GGS victory circle permutations = 4P3 = 24 GGG victory circle permutations = 4P3 = 24

Total possible permutations of victory circle = 24*4 = 96 (Answer C)?

Thanks for the clarification - I'm still not comfortable with the answer and i have three more questions:

1) Is a "victory circle" an actual circle and therefore are there rotational symmetry considerations that need to be taken into account when calculating arrangements. [I've never heard of this term before - so apologies if this is a silly question?]

2) The question asks: "three medal winners stand together with their medals to form a victory circle" so are we calculating permutations with to many people in the 'victory circle'? [This is the only flaw i'm pretty sure about which takes the number of permutation down to 13 - which is not an answer]

3)The questions asks: "the three medal winners stand together with their medals" so should we be calculating permutations of people and medals and not discounting duplicate medals? [Potentially tenuous but needed to get to an answer - see below]

Assuming that (1) a 'victory circle' is not a circle but a straight line, (2) that we are only concerned with medal holders, and (3) person A with a silver medal can be considered different to person B with a silver medal (in a scenario where they both have one) then:

Possible medal combinations: GSB, GSS, GGS, GGG

GSB victory circle permutations = 4P3 = 24 GSS victory circle permutations = 4P3 = 24 GGS victory circle permutations = 4P3 = 24 GGG victory circle permutations = 4P3 = 24

Total possible permutations of victory circle = 24*4 = 96 (Answer C)?

EDIT: for clarity

OK, first of all victory circle is not the actual circle. Question asks how many different scenarios are possible for: medal-person; medal-person; medal-person.

Let's consider the easiest one - scenario GGG. You say there are 24 circles (scenarios) possible, but there are only 4:

1. G/a, G/b, G/c (here G/b, G/a, G/c is the same scenario a, b and c won the gold); 2. G/a, G/b, G/d; 3. G/a, G/c, G/d; 4. G/b, G/c, G/d.

Scenario GGS: 1. G/a, G/b, S/c (a and b won gold and c won silver. Here G/b, G/a, S/c is the same scenario: a and b won gold and c won silver; 2. G/a, G/b, S/d; 3. G/a, G/c, S/b; 4. G/a, G/c, S/d; 5. G/a, G/d, S/b; 6. G/a, G/d, S/c;

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

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20 Dec 2014, 10:37

If this question comes, you are already at QA 60.

4C3 = 4 ways of selecting 3 people.

G S B 3 0 0 -> only 1 way as 3 identical gold medals are given to all the three people 2 1 0 -> 3 ways as GGS, GSG, SGG 1 2 0 -> 3 ways 1 1 1 -> 3! = 6 ways

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

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20 Dec 2014, 22:18

Bunuel wrote:

Ramsay wrote:

Quote:

Possible scenarios are:

1. Gold/Silver/Bronze/No medal (no ties) - 4!=24; 2. Gold/Gold/Silver/No medal - 4!/2!=12; 3. Gold/Silver/Silver/No medal - 4!/2!=12; 4. Gold/Gold/Gold/No medal - 4!/3!=4.

Total: 24+12+12+4=52

Answer: B.

Sorry, but i dont see how this answers the "how many different victory circles are possible" part of the question?

If we are concerned only with medals in the victory circle:

Possible medal combinations are: GSB, GSS, GGS, GGG. Possible victory circles [N.B. removing rotationally symmetric permutations]:

GSB: = 2 GSS: = 1 GGS: = 1 GGG: = 1

So, total possible victory circles are 5.

If we are concerned with combinations of people and medals in the victory circle:

Possible permutation of winners (top 3) = 4P3 = 24 Possible victory circles:

Please can i get some advice on where i went wrong with both of these approaches?

I don't quite understand your solution. Here is the logic behind mine:

We have four possible patterns (GSBN, GSSN, GGSN, GGGN) and four persons (a, b, c, d). Each victory circle is made by assigning pattern letters to these persons as follows: Firs pattern: GSBN a--b--c--d G--S--B--N G--B--S--N G--B--N--S ... So how many victory circles are possible for the first pattern? It would be the # of permutations of the letters GSBN, as these are four distinct letters, the # is 4!=24.

The same for other patterns: Second pattern: GSSN. # of permutations is 4!/2!=12 (as there are four letters out of which 2 are the same).

Third pattern: GGSN. # of permutations is 4!/2!=12.

Fourth pattern: GGGN. # of permutations is 4!/3!=4. a--b--c--d G--G--G--N G--G--N--G G--N--G--G N--G--G--G Here you can see that these four victory circles are all different and the same will be for other patterns.

24+12+12+4=52

Hope it's clear.

RaviChandra wrote:

what if 2 people come first(i.e. Gold) and another 2 come second position(silver)

We are told in the stem that only 3 medals will be awarded, so we should take this as a fact.

Question: 1. Why do we take 'N' into consideration? why cant the patterns be : GGG, GGS, GSS, GSB? 2. considering patterns mentioned in your above explanation, why isnt GSBB a pattern too ?

Question: 1. Why do we take 'N' into consideration? why cant the patterns be : GGG, GGS, GSS, GSB? 2. considering patterns mentioned in your above explanation, why isnt GSBB a pattern too ?

1. We have 4 people in the race. 2. We are told that exactly three medals are awarded not 4. _________________

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

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21 Dec 2014, 06:39

Bunuel wrote:

vinbitstarter wrote:

Question: 1. Why do we take 'N' into consideration? why cant the patterns be : GGG, GGS, GSS, GSB? 2. considering patterns mentioned in your above explanation, why isnt GSBB a pattern too ?

1. We have 4 people in the race. 2. We are told that exactly three medals are awarded not 4.

Thank you very much!!

gmatclubot

Re: In a 4 person race, medals are awarded to the fastest 3 runn
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21 Dec 2014, 06:39

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