Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 22 Jul 2008
Posts: 96
Location: Bangalore,Karnataka

In a 4 person race, medals are awarded to the fastest 3 runn [#permalink]
Show Tags
03 Dec 2009, 07:52
1
This post received KUDOS
8
This post was BOOKMARKED
Question Stats:
33% (02:40) correct
67% (02:03) wrong based on 157 sessions
HideShow timer Statistics
In a 4 person race, medals are awarded to the fastest 3 runners. The firstplace runner receives a gold medal, the secondplace runner receives a silver medal, and the thirdplace 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 twoway tie for firstplace, the top two runners receive gold medals, the nextfastest 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
Official Answer and Stats are available only to registered users. Register/ Login.



Math Expert
Joined: 02 Sep 2009
Posts: 39626

Re: race [#permalink]
Show Tags
03 Dec 2009, 10:42
2
This post received KUDOS
Expert's post
4
This post was BOOKMARKED
kirankp wrote: In a 4 person race, medals are awarded to the fastest 3 runners. The firstplace runner receives a gold medal, the secondplace runner receives a silver medal, and the thirdplace 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 twoway tie for firstplace, the top two runners receive gold medals, the nextfastest 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. Total: 24+12+12+4=52 Answer: B.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 31 Mar 2010
Posts: 17
Schools: Stanford GSB (waiting), Wharton (wl w/ int), INSEAD (admit)
WE 1: TopTier MC

Re: race [#permalink]
Show Tags
06 Apr 2010, 03:44
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: GSB: = 24*2 = 48 GSS: = 48 [N.B. S1 <> S2] GGS: = 48 GGG: = 48 So, total possible victory circles are 192... Please can i get some advice on where i went wrong with both of these approaches?



BSchool Forum Moderator
Joined: 02 Oct 2009
Posts: 592
GMAT 1: 530 Q47 V17 GMAT 2: 710 Q50 V36
WE: Business Development (Telecommunications)

Re: race [#permalink]
Show Tags
06 Apr 2010, 23:31
what if 2 people come first(i.e. Gold) and another 2 come second position(silver)



Math Expert
Joined: 02 Sep 2009
Posts: 39626

Re: race [#permalink]
Show Tags
07 Apr 2010, 01:45
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: GSB: = 24*2 = 48 GSS: = 48 [N.B. S1 <> S2] GGS: = 48 GGG: = 48 So, total possible victory circles are 192... 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 abcdGSBN GBSN GBNS ... 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. abcdGGGN GGNG GNGG NGGG 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.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 31 Mar 2010
Posts: 17
Schools: Stanford GSB (waiting), Wharton (wl w/ int), INSEAD (admit)
WE 1: TopTier MC

Re: race [#permalink]
Show Tags
07 Apr 2010, 02:30
Bunuel, 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



Math Expert
Joined: 02 Sep 2009
Posts: 39626

Re: race [#permalink]
Show Tags
07 Apr 2010, 03:41
Ramsay wrote: Bunuel, 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: medalperson; medalperson; medalperson. 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; 7. G/b, G/c, S/a; 8. G/b, G/c, S/d; 9. G/b, G/d, S/a; 10. G/b, G/d, S/c; 11. G/c, G/d, S/a; 12. G/c, G/d, S/b. You can see that all 12 scenarios are different in terms of medal/person, medal/person, medal/person. The same will be for GSS. For GSB you are right there will be 4P3=24 different scenarios, or as I wrote 4!=24. So again 24(GSB)+12(GGS)+12(GSS)+4(GGG)=52 (which is OA according to kirankp). Hope it's clear.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 31 Mar 2010
Posts: 17
Schools: Stanford GSB (waiting), Wharton (wl w/ int), INSEAD (admit)
WE 1: TopTier MC

Re: race [#permalink]
Show Tags
07 Apr 2010, 06:59
Thanks! I didn't understand the approach end to end but that's a fantastic explanation. +1



Manager
Joined: 07 Jan 2010
Posts: 240

Re: race [#permalink]
Show Tags
09 Apr 2010, 23:58
super explanation ...



Intern
Joined: 06 Oct 2013
Posts: 3

Re: In a 4 person race, medals are awarded to the fastest 3 runn [#permalink]
Show Tags
16 Aug 2014, 10:00
I get that there are 52 arrangements of medals and people, but
if they stand in a circle of 3 people, there are 2 circles that can be formed per arrangement above
so the total number of possible victory circles is 104
What do you think?



Current Student
Joined: 03 Feb 2013
Posts: 941
Location: India
Concentration: Operations, Strategy
GPA: 3.88
WE: Engineering (Computer Software)

Re: In a 4 person race, medals are awarded to the fastest 3 runn [#permalink]
Show Tags
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 So 1 + 6 + 6 = 13 Hence 13*4 = 52 B)
_________________
Thanks, Kinjal My Debrief : http://gmatclub.com/forum/hardworknevergetsunrewardedforever189267.html#p1449379 My Application Experience : http://gmatclub.com/forum/hardworknevergetsunrewardedforever18926740.html#p1516961 Linkedin : https://www.linkedin.com/in/kinjaldas/
Please click on Kudos, if you think the post is helpful



Intern
Joined: 07 Oct 2014
Posts: 14

Re: In a 4 person race, medals are awarded to the fastest 3 runn [#permalink]
Show Tags
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: GSB: = 24*2 = 48 GSS: = 48 [N.B. S1 <> S2] GGS: = 48 GGG: = 48 So, total possible victory circles are 192... 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 abcdGSBN GBSN GBNS ... 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. abcdGGGN GGNG GNGG NGGG 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 ?



Math Expert
Joined: 02 Sep 2009
Posts: 39626

Re: In a 4 person race, medals are awarded to the fastest 3 runn [#permalink]
Show Tags
21 Dec 2014, 04:27



Intern
Joined: 07 Oct 2014
Posts: 14

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



Intern
Joined: 26 Jul 2016
Posts: 23
Concentration: Finance, Strategy
WE: Analyst (Energy and Utilities)

In a 4 person race, medals are awarded to the fastest 3 runn [#permalink]
Show Tags
14 Oct 2016, 23:58
1
This post received KUDOS
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 2WAY tie? If there is a 2WAY tie for FIRST, then the medals awarded are: GOLD, GOLD, SILVER. If there is a 2WAY tie for SECOND, then the medals awarded are: GOLD, SILVER, SILVER. There cannot be a 2WAY tie for THIRD (because exactly three medals are awarded intotal). (3) What if there is a 3WAY tie? If there is a 3WAY tie for FIRST, then the medals awarded are: GOLD, GOLD, GOLD. There are no other possible 3WAY 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 GoldGoldSilver 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 GoldGoldGold 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: AlbertGOLD, CamiGOLD, BobGOLD. BobGOLD, AlbertGOLD, CamiGOLD. BobGOLD, CamiGOLD, AlbertGOLD. CamiGOLD, AlbertGOLD, BobGOLD. CamiGOLD, BobGOLD, AlbertGOLD. 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 GOLDSILVERBRONZE victory circles. (Combination 2) 12 unique GOLDGOLDSILVER victory circles. (Combination 3) 12 unique GOLDSILVERSILVER victory circles. (Combination 4) 4 unique GOLDGOLDGOLD 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



Senior Manager
Joined: 26 Oct 2016
Posts: 461
Location: United States
Concentration: Marketing, International Business
GPA: 4
WE: Education (Education)

Re: In a 4 person race, medals are awarded to the fastest 3 runn [#permalink]
Show Tags
03 Jan 2017, 21:21
Bunuel wrote: kirankp wrote: In a 4 person race, medals are awarded to the fastest 3 runners. The firstplace runner receives a gold medal, the secondplace runner receives a silver medal, and the thirdplace 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 twoway tie for firstplace, the top two runners receive gold medals, the nextfastest 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. Total: 24+12+12+4=52 Answer: B. I understand complete solution except this part "4. Gold/Gold/Gold/No medal  4!/3!=4." Why we are considering this case?
_________________
Thanks & Regards, Anaira Mitch



Math Expert
Joined: 02 Sep 2009
Posts: 39626

Re: In a 4 person race, medals are awarded to the fastest 3 runn [#permalink]
Show Tags
04 Jan 2017, 00:55
anairamitch1804 wrote: Bunuel wrote: kirankp wrote: In a 4 person race, medals are awarded to the fastest 3 runners. The firstplace runner receives a gold medal, the secondplace runner receives a silver medal, and the thirdplace 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 twoway tie for firstplace, the top two runners receive gold medals, the nextfastest 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. Total: 24+12+12+4=52 Answer: B. I understand complete solution except this part "4. Gold/Gold/Gold/No medal  4!/3!=4." Why we are considering this case? This is a case when 3 runners are tied for gold medal.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Director
Joined: 14 Nov 2016
Posts: 974
Location: Malaysia

In a 4 person race, medals are awarded to the fastest 3 runn [#permalink]
Show Tags
30 Mar 2017, 22:36
1
This post received KUDOS
kirankp wrote: In a 4 person race, medals are awarded to the fastest 3 runners. The firstplace runner receives a gold medal, the secondplace runner receives a silver medal, and the thirdplace 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 twoway tie for firstplace, the top two runners receive gold medals, the nextfastest 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 OFFICIAL SOLUTION 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 2WAY tie? If there is a 2WAY tie for FIRST, then the medals awarded are: GOLD, GOLD, SILVER. If there is a 2WAY tie for SECOND, then the medals awarded are: GOLD, SILVER, SILVER. There cannot be a 2WAY tie for THIRD (because exactly three medals are awarded in total). (3) What if there is a 3WAY tie? If there is a 3WAY tie for FIRST, then the medals awarded are: GOLD, GOLD, GOLD. There are no other possible 3WAY 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, BronzeGold 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 \(4*3*2=24\) 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 GoldGoldSilver 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 GoldGoldGold 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: AlbertGOLD, CamiGOLD, BobGOLD. BobGOLD, AlbertGOLD, CamiGOLD. BobGOLD, CamiGOLD, AlbertGOLD. CamiGOLD, AlbertGOLD, BobGOLD. CamiGOLD, BobGOLD, AlbertGOLD. 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 \(\frac{24}{6}=4\) unique victory circles that contain 3 GOLD medalists. FINALLY, then, we have the following: (Combination 1) 24 unique GOLDSILVERBRONZE victory circles. (Combination 2) 12 unique GOLDGOLDSILVER victory circles. (Combination 3) 12 unique GOLDSILVERSILVER victory circles. (Combination 4) 4 unique GOLDGOLDGOLD victory circles. Thus, there are \(24+12+12+4=52\) unique victory circles. The correct answer is B. Bunuel, Original source is Manhattan Prep. Please help to edit. Thank you.
Attachments
Untitled.jpg [ 74.4 KiB  Viewed 448 times ]
_________________
"Be challenged at EVERY MOMENT."
“Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.”
"Each stage of the journey is crucial to attaining new heights of knowledge."
Rules for posting in verbal forum  Please DO NOT post short answer in your post!



Math Expert
Joined: 02 Sep 2009
Posts: 39626

Re: In a 4 person race, medals are awarded to the fastest 3 runn [#permalink]
Show Tags
30 Mar 2017, 23:29
ziyuen wrote: kirankp wrote: In a 4 person race, medals are awarded to the fastest 3 runners. The firstplace runner receives a gold medal, the secondplace runner receives a silver medal, and the thirdplace 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 twoway tie for firstplace, the top two runners receive gold medals, the nextfastest 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 Bunuel, Original source is Manhattan Prep. Please help to edit. Thank you. Edited. Thank you.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 27 Dec 2015
Posts: 7

Re: In a 4 person race, medals are awarded to the fastest 3 runn [#permalink]
Show Tags
05 May 2017, 09:48
One of the key things mentioned here is " they stand in a victory circle" so shouldn't we multiple 52 with 2!, for making an appropriate circle arrangement in each case.




Re: In a 4 person race, medals are awarded to the fastest 3 runn
[#permalink]
05 May 2017, 09:48



Go to page
1 2
Next
[ 21 posts ]




