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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, 06:52

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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, 09:37

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, 21:18

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, 05:39

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.

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

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14 Oct 2016, 22:58

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I am posting both wordy but useful explanation ( just for the understanding ) and shorter version.

Longer version:

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 intotal). (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-GoldSilver 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 12unique 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 GoldGold-Gold victory circles: Albert is awarded a GOLD. Bob is awarded a GOLD. Cami is awarded a GOLD. Notice that this victory circleis exactly the same as the following victory circles: Albert-GOLD, CamiGOLD, 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 circlehas 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 circlesthat contain 3 GOLD medalists. FINALLY, then, we have the following:

(Combination 1) 24 unique GOLD-SILVER-BRONZE victory circles. (Combination 2) 12 unique GOLDGOLD-SILVER victory circles. (Combination 3) 12 unique GOLD-SILVER-SILVER victory circles. (Combination 4) 4 unique GOLD-GOLD-GOLD victory circles. Thus, there are uniquevictory circles. The correct answer is B.

Shorter version:

Case 1: G+S+B --> 4P3 = 4C3 3! = 24 Case 2: G+G+S --> 4C2 * 2C1 = 12 Case 3: G+S+S --> 4C2 * 2C1 = 12 Case 4: G+G+G --> 4C3 = 4

24+12+12+4=52

All the best

gmatclubot

In a 4 person race, medals are awarded to the fastest 3 runn
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14 Oct 2016, 22:58

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