Find all School-related info fast with the new School-Specific MBA Forum

It is currently 30 Sep 2014, 09:54

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

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

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Senior Manager
Senior Manager
avatar
Joined: 05 Jan 2006
Posts: 383
Followers: 1

Kudos [?]: 18 [0], given: 0

In a 4 person race, medals are awarded to the fastest 3 [#permalink] New post 11 Apr 2006, 22:13
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
Senior Manager
Senior Manager
avatar
Joined: 05 Jan 2006
Posts: 383
Followers: 1

Kudos [?]: 18 [0], given: 0

 [#permalink] New post 11 Apr 2006, 22:15
This is One old MGMT Challage!

OA is B and I have no idea how to get this! I tried what ever I can think off! That includes reading explaination! but still I don't get it!
Senior Manager
Senior Manager
avatar
Joined: 22 Nov 2005
Posts: 482
Followers: 1

Kudos [?]: 3 [0], given: 0

 [#permalink] New post 12 Apr 2006, 12:49
chiragr wrote:
This is One old MGMT Challage!

OA is B and I have no idea how to get this! I tried what ever I can think off! That includes reading explaination! but still I don't get it!
\


Can you put explanation here
Manager
Manager
avatar
Joined: 23 Jan 2006
Posts: 193
Followers: 1

Kudos [?]: 2 [0], given: 0

 [#permalink] New post 12 Apr 2006, 12:57
hmmmm....
I'm going to pick A.
Not very mathematical, but-

Out of three winners - there are 6 possible permutations (think trifecta @ the track ;))

and there are 4 ways (3C4) to choose those three winners.

6*4 = 24

So I guess the "mathematical" way is 3P3 * 3C4 = 3! * 4!/(3!*1!) = 24
Manager
Manager
avatar
Joined: 23 Jan 2006
Posts: 193
Followers: 1

Kudos [?]: 2 [0], given: 0

 [#permalink] New post 12 Apr 2006, 13:04
nevermind - i read too fast - didn't count the "ties" in there...
VP
VP
avatar
Joined: 28 Mar 2006
Posts: 1389
Followers: 2

Kudos [?]: 18 [0], given: 0

 [#permalink] New post 12 Apr 2006, 15:03
Heres how I tried.

There are 4 possible ways in which the 3 positions for awards can be done

F S T
F F S
F S S
F F F

(F- First position S Second position and T third position)

we have 4 players so we have to select 3 from 4 so 4C3 ways = 4 --(i)

Now come to the arrangements


F S T 3! = 6 ways
F F S and F S S in 3!/2! ways for each one which is 3+3 = 6 ways
F F F in 1 way 3!/3! = 1

So 6+6+1 = 13

Multiplied by (i) which is the selections = 13*4 =52 ways....
Senior Manager
Senior Manager
avatar
Joined: 05 Jan 2006
Posts: 383
Followers: 1

Kudos [?]: 18 [0], given: 0

 [#permalink] New post 12 Apr 2006, 15:24
gmat_crack wrote:
chiragr wrote:
This is One old MGMT Challage!

OA is B and I have no idea how to get this! I tried what ever I can think off! That includes reading explaination! but still I don't get it!
\


Can you put explanation here


I am waiting for few more to reply..
Senior Manager
Senior Manager
avatar
Joined: 05 Jan 2006
Posts: 383
Followers: 1

Kudos [?]: 18 [0], given: 0

 [#permalink] New post 12 Apr 2006, 23:31
Answer
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 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-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 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 unique victory circles.

The correct answer is B.
  [#permalink] 12 Apr 2006, 23:31
    Similar topics Author Replies Last post
Similar
Topics:
2 Experts publish their posts in the topic Medals are to be awarded to three teams in a 10-team anilnandyala 8 02 Oct 2010, 02:54
4 Experts publish their posts in the topic In a 4 person race, medals are awarded to the fastest 3 runn kirankp 9 03 Dec 2009, 06:52
3 Experts publish their posts in the topic In a 4 person race, medals are awarded to the fastest 3 runn rathoreaditya81 6 20 Nov 2009, 01:53
Medals TheRob 3 17 Jun 2009, 05:23
21 Experts publish their posts in the topic In a 4 person race, medals are awarded to the fastest 3 runn 12345678 10 18 Sep 2007, 22:32
Display posts from previous: Sort by

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

  Question banks Downloads My Bookmarks Reviews Important topics  


cron

GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.