In a 4 person race, medals are awarded to the fastest 3 runners. The

Question

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

A. 24
B. 52
C. 96
D. 144
E. 648

Bunuel · Accepted Answer

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

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.

powerka · Answer

This is a Manhattan GMAT question. OA is B.

It can be found on "Manhattan GMAT Archive (tough problems set).doc" and on https://www.manhattangmat.com/forums/gma ... -t886.html.

I don't like their answer; too Manhattan like = too wordy.

Here's my answer:

Scenario 1: g+s+b --> 4P3 = 4C3 3! = 24
Scenario 2: g+g+s --> 4C2 * 2C1 = 12
Scenario 3: g+s+s --> 4C2 * 2C1 = 12
Scenario 4: g+g+g --> 4C3 = 4

24+12+12+4=52

This answer assumes that the order of the runners receiving same medals, which is random, does not affect the positioning on the victory circle.

Ramsay · Answer

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 : 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 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?

Bunuel · Answer

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.

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

Ramsay · Answer

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

Bunuel · Answer

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;

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.

Ramsay · Answer

Thanks! I didn't understand the approach end to end but that's a fantastic explanation. +1

Indien · Answer

super explanation ...

KarishmaB · Answer

A victory circle is where winners stand together in any formation. The reason victory circle is used here is that the question does not ask you in how many ways can you have different winners. It asks you in how many ways can you have different winners or different medals. e.g. if there are 4 runners A, B, C and D, and if A, B and C are winners, A - Gold, B - Silver, C - Bronze is different from A - Gold, B - Gold, C - Gold. Hence they have included runners as well as their medals in the victory circle. In how many ways can medals be awarded? Ques3.jpg There are 4 ways: GGG - Select 3 of the 4 people - 4C3 = 4 GGS - Select 3 people and arrange the medals among them - 4C3 * 3!/2! = 12 GSS - Select 3 people and arrange the medals among them - 4C3 * 3!/2! = 12 GSB - Select 3 people and arrange the medals among them - 4C3 * 3! = 24 Total different victory circles = 4+12+12+24 = 52

vinstargazer · Answer

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 ?

Bunuel · Answer

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

deekshant111 · Answer

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

anairamitch1804 · Answer

I understand complete solution except this part "4. Gold/Gold/Gold/No medal - 4!/3!=4." Why we are considering this case?

Bunuel · Answer

This is a case when 3 runners are tied for gold medal.

BillyZ · Answer

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 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 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 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 \frac{24}{6}=4 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 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. :-D

IIMC · Answer

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.

Bunuel · Answer

This is addressed above: victory circle is not the actual circle. Question asks how many  different  scenarios are possible for: medal-person; medal-person; medal-person.

streetking · Answer

fantastic explanation

i missed the all Golds circle

Fdambro294 · Answer

Ok....

So “victory circle” does NOT mean we are taking account of the different circular arrangements that are possible. We are just determining who is in the victory circle and who is out of the victory circle. Come to think of it, does Manhattan Prep even touch the concept of circular arrangements?

It’s as if we are distributing medals (sometimes identical) to ———-> different groups of 3 different people out of 4 total people. How many different distributions can we have?

Once you get through the issue of “victory circle” and realize that the concept of circular arrangements does not come into play, it becomes a more straightforward problem.

4 people- A , B , C , D

Scenario 1: there are no ties and we have: Gold - Silver - Bronze or G-S-B

(1st) determine how many different ways we can choose 3 people out of 4 to be in the group that receives the medals:

“4 choose 3” = 4! / (3! 1!) = 4 ways

And

For each possible combination, the medals can be arranged 3! Ways

4 * 3! = 24 ways

Scenario 2: Gold - Gold - Silver or G-G-S (two way tie for 1st)

(1st) how many ways we can have 3 of the 4 runners receive one of the medals

“4 choose 3” = 4! / (3! 1!) = 4 ways

AND

(2nd)How many different ways are there to arrange 3 items (2 of which are identical) among each of the groups of 3 people that are possible

3! / 2! = 3

4 * 3 = 12 ways

Scenario 3: tied for 2nd place - have G - S - S

Same exact logic as scenario 2 = 12 ways

Scenario 4: have all 3 runners tie for 1st place (they would all receive the same medal) B - B - B (or gold gold gold, doesn’t matter as long as you realize they all receive the same medal.

(1st)the number of ways we can have 3 of the 4 runners chosen to receive a medal and appear in the victory circle

“4 choose 3” = 4! / (3! 1!) = 4 ways

AND

(2nd) for each possible group of 3 chosen runners, there will be only 1 way for the runners to receive the medals because they will all receive the same one

1

4 * 1 = 4 ways

Add up the 4 scenarios:

24 + 12 + 12 + 4 = 52

52 ways

Another thing to understand in some of the solutions is why they included “no medal”.

When you choose which 3 people receive the medal, you are automatically choosing which person does NOT receive the medal.

So you can answer the question by setting up the 4 slots (for Ex: G - S - B - No medal) and then arranging the 4 people among the slots.

All I did above was arrange the medals among the people chosen (but you have to make sure to account for the different possibilities of 3 people that can be chosen...and make sure each person has an equal chance of being included in any given arrangement).

That’s why I determined all the different groups that are possible first, and then for each different group of 3, I arranged the medals among those chosen peope.

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