GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 23 Oct 2019, 16:26

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.

Close

Request Expert Reply

Confirm Cancel

Coach Miller is filling out the starting lineup for his indoor soccer

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Manager
Manager
avatar
Joined: 19 Aug 2007
Posts: 148
Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

Show Tags

New post Updated on: 29 Apr 2019, 22:42
2
21
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

64% (02:36) correct 36% (02:58) wrong based on 344 sessions

HideShow timer Statistics

Coach Miller is filling out the starting lineup for his indoor soccer team. There are 10 boys on th team, and he must assign 6 starters to the following positions: 1 goalkeeper, 2 on defense, 2 in midfield, and 1 forward. Only 2 of the boys can play goalkeeper, and they cannot play any other positions. The other boys can each play any of the other positions. How many different groupings are possible?

A. 60
B. 210
C. 2580
D. 3360
E. 151200

Originally posted by jimjohn on 25 Dec 2007, 16:44.
Last edited by Bunuel on 29 Apr 2019, 22:42, edited 3 times in total.
Renamed the topic and edited the question.
Most Helpful Expert Reply
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58464
Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

Show Tags

New post 25 Oct 2009, 18:08
5
1
rvthryet wrote:
A Coach is filling out the starting lineup for his indoor soccer team. There are 10 boys on the team, and he must assign 6 starters to the following positions: 1 goalkeeper, 2 on defence, 2 in midfield, and 1 forward. Only 2 of the boys can play goalkeeper, and they cannot play any other positions. The other boys can each play any of the other positions. How many different groupings are possible?

A) 60

B) 210

C) 2580

D) 3360

E) 151200


2C1 select 1 goalkeeper from 2 boys;
8C2 select 2 defense from 8 boys (as 2 boys can only play goalkeeper 10-2=8);
6C2 select 2 midfield from 6 boys (as 2 boys can only play goalkeeper and 2 we've already selected for defense 10-2-2=6);
4C1 select 1 forward from 4 boys (again as 2 boys can play only goalkeeper, 4 we've already selected for defense and midfield 10-2-4=4)

Total # of selection=2C1*8C2*6C2*4C1=3360

Answer: D.
_________________
Most Helpful Community Reply
SVP
SVP
User avatar
Joined: 21 Jan 2007
Posts: 2240
Location: New York City
Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

Show Tags

New post 25 Dec 2007, 17:48
4
1
I hope this is NOT a GMATprep question....


There are 4 slots.

_ _ _ _

2C1 * 8C2 * 6C2 * 4C1
= 2 * 28 * 15 * 4
= 56 * 60
= 3360

Recognize that for the second slot, we only have 10-2 = 8 elements to choose from. We need 2 of 8 elements to fill that spot. 8C2
For the third slot, we only have 6 elements left to choose from. We need to fill it with 2 elements.
General Discussion
CEO
CEO
User avatar
B
Joined: 17 Nov 2007
Posts: 3038
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
GMAT ToolKit User
Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

Show Tags

New post 25 Dec 2007, 19:18
jimjohn wrote:
Coach Miller is filling out the starting lineup for his indoor soccer team. There are 10 boys on th team, and he must assign 6 starters to the following positions: 1 goalkeeper, 2 on defense, 2 in midfield, and 1 forward. Only 2 of the boys can play goalkeeper, and they cannot play any other positions. The other boys can each play any of the other positions. How many different groupings are possible?

A) 60

B) 210

C) 2580

D) 3360

E) 151200


N=8P5*2P1/(2P2*2P2)=3360
VP
VP
avatar
Joined: 04 May 2006
Posts: 1368
Schools: CBS, Kellogg
Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

Show Tags

New post 05 Jan 2008, 21:04
walker wrote:
N=8P5*2P1/(2P2*2P2)=3360


Walker,

I cant get it. Why it is not C, but P here or your resulting calculation is random, pl give yr logic explaination!
_________________
CEO
CEO
User avatar
B
Joined: 17 Nov 2007
Posts: 3038
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
GMAT ToolKit User
Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

Show Tags

New post 06 Jan 2008, 00:27
3
1
sondenso wrote:
I cant get it. Why it is not C, but P here or your resulting calculation is random, pl give yr logic explaination!


N=8P5*2P1/(2P2*2P2)=3360
8P5 - we choose 5 boys of 8 (without goalkeepers) for 5 positions: 2 on defense, 2 in midfield, and 1 forward.
2P2*2P2 - we can change position within 2 on defense, 2 in midfield. So, we should exclude this variations.
2P1=2C1 - we choose goalkeeper of 2 boys
_________________
HOT! GMAT Club Forum 2020 | GMAT ToolKit 2 (iOS) - The OFFICIAL GMAT CLUB PREP APPs, must-have apps especially if you aim at 700+ | Limited Online GMAT/GRE Math tutoring
Intern
Intern
avatar
Joined: 02 Oct 2009
Posts: 15
Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

Show Tags

New post 25 Oct 2009, 18:57
1
Small doubt.. Why should this not be 2C1 x 8C5??

I just can't seem to understand how is my thinking flawed there, although it is quite obvious that it is :oops:
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58464
Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

Show Tags

New post 25 Oct 2009, 19:38
5
1
rvthryet wrote:
Small doubt.. Why should this not be 2C1 x 8C5??

I just can't seem to understand how is my thinking flawed there, although it is quite obvious that it is :oops:


Imagen different situation 4 players, we should choose 1 for defense and 1 for forward. (no restrictions).

The way you are doing you'll get 4C2=6. But look at the real case.

ABCD (players):

Defence - Forward
A B
A C
A D

B A
B C
B D

C A
C B
C D

D A
D B
D C

Total 12 possibilities 4C1*3C1=4*3=12. You just narrowed possible ways of selection.

In original question we are not choosing 5 people from 8, but we are choosing 2 from 8, than 2 from 6, than 1 from 4 (well and before we chose 1 from 2 as goalkeeper). And this is more ways of selection than 8C5 as you can see in the example.
_________________
Manager
Manager
avatar
Joined: 15 Sep 2009
Posts: 90
Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

Show Tags

New post 26 Oct 2009, 07:12
1
2 goal keepers can be selected in 2 ways.

Rest 5 positions has to be filled from 8 boys.

2 Defence can be selected in 8C2 ways

2 Midfielder can be selected in 6C2 ways

1 forward can be selected in 4C1 ways.

So the total combinations 8*28*15*4 = 3360 ways
Manager
Manager
avatar
Status: MBA Aspirant
Joined: 12 Jun 2010
Posts: 120
Location: India
Concentration: Finance, International Business
WE: Information Technology (Investment Banking)
Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

Show Tags

New post 02 Feb 2012, 21:30
Bunuel wrote:
rvthryet wrote:
A Coach is filling out the starting lineup for his indoor soccer team. There are 10 boys on the team, and he must assign 6 starters to the following positions: 1 goalkeeper, 2 on defence, 2 in midfield, and 1 forward. Only 2 of the boys can play goalkeeper, and they cannot play any other positions. The other boys can each play any of the other positions. How many different groupings are possible?

A) 60

B) 210

C) 2580

D) 3360

E) 151200


2C1 select 1 goalkeeper from 2 boys;
8C2 select 2 defense from 8 boys (as 2 boys can only play goalkeeper 10-2=8);
6C2 select 2 midfield from 6 boys (as 2 boys can only play goalkeeper and 2 we've already selected for defense 10-2-2=6);
4C1 select 1 forward from 4 boys (again as 2 boys can play only goalkeeper, 4 we've already selected for defense and midfield 10-2-4=4)

Total # of selection=2C1*8C2*6C2*4C1=3360

Answer: D.

Bunnel for these type of question do we need to follow the positions as given in the question stem. Cant we first select 1 forward first and then the defence and midfield. If we go this way the no of selections will become 2c1*8c1*7c2*5c2. Lemme know your views on this
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58464
Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

Show Tags

New post 02 Feb 2012, 23:47
subhajeet wrote:
Bunnel for these type of question do we need to follow the positions as given in the question stem. Cant we first select 1 forward first and then the defence and midfield. If we go this way the no of selections will become 2c1*8c1*7c2*5c2. Lemme know your views on this


You can choose 2 in defense, 2 in midfield, and 1 forward from 8 in ANY order, you'l get the same result. The formula gives # of different selections of 2, 2, and 1 possible from 8, so the result must be the same despite the order in which you make this selections.
_________________
Senior Manager
Senior Manager
User avatar
Joined: 23 Mar 2011
Posts: 398
Location: India
GPA: 2.5
WE: Operations (Hospitality and Tourism)
Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

Show Tags

New post 05 Feb 2012, 12:08
Hi Bunuel, I am not clear as to do this with P or C...though your answer with C matches mine with P. My explanation is same as Walkers. except for goalkeepers all other positions are common - should we not then divide by 2P2*2P2? Since selecting 2 goalkeepers was critical in terms of order - i used P.pls advice
_________________
"When the going gets tough, the tough gets going!"

Bring ON SOME KUDOS MATES+++



-----------------------------
Quant Notes consolidated: http://gmatclub.com/forum/consolodited-quant-guides-of-forum-most-helpful-in-preps-151067.html#p1217652

My GMAT journey begins: http://gmatclub.com/forum/my-gmat-journey-begins-122251.html

All about Richard Ivey: http://gmatclub.com/forum/all-about-richard-ivey-148594.html#p1190518
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58464
Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

Show Tags

New post 05 Feb 2012, 12:31
sdas wrote:
Hi Bunuel, I am not clear as to do this with P or C...though your answer with C matches mine with P. My explanation is same as Walkers. except for goalkeepers all other positions are common - should we not then divide by 2P2*2P2? Since selecting 2 goalkeepers was critical in terms of order - i used P.pls advice


P and C just represent different formulas, different ways of counting. Most combinations questions can be solved in multiple ways, and if you understand the concept it really doesn't matter which approach you take.

As for this question: since we are dealing with different groups to be chosen from total and the order in each specific group doesn't matter I would use the method described in my previous post. It seems more straightforward and easy, at least for me.
_________________
Intern
Intern
User avatar
Joined: 12 Oct 2012
Posts: 9
Location: Singapore
Concentration: International Business, General Management
GMAT 1: 710 Q49 V35
GPA: 3.65
Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

Show Tags

New post 05 Nov 2012, 21:07
2
1
The formula can be simplified for slot method.

First is goalkeeper restriction: 2 options.
Next: Out of rest 8 players we need to fill slots with 2, 2, 1 and 3 (non-assigned) players.
Numbers of people in the same position are listed in denumeretor with factorial.

8! / (2! x 2! x 1! x 3!) = 1680
2 x 1680 = 3360.

As you can see the order of picking players doesnt matter.

Solution of Bunuel simply leads to the same formula:
\(2 * \frac{8!}{2!6!} * \frac{6!}{2!4!} * \frac{4!}{1!3!} = 2 * \frac{8!}{2!2!3!}\)
Senior Manager
Senior Manager
User avatar
Joined: 13 Aug 2012
Posts: 399
Concentration: Marketing, Finance
GPA: 3.23
GMAT ToolKit User
Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

Show Tags

New post 28 Dec 2012, 04:45
2
rvthryet wrote:
A Coach is filling out the starting lineup for his indoor soccer team. There are 10 boys on the team, and he must assign 6 starters to the following positions: 1 goalkeeper, 2 on defence, 2 in midfield, and 1 forward. Only 2 of the boys can play goalkeeper, and they cannot play any other positions. The other boys can each play any of the other positions. How many different groupings are possible?

A. 60
B. 210
C. 2580
D. 3360
E. 151200


How many ways to select goal keeper? 2 only, others cannot do it
How many ways to select 2 in midfield? \(=\frac{8!}{2!6!}=28\)
How many ways to select 2 on defence? \(=\frac{6!}{2!4!}=15\)
How many ways to select 1 in forward? \(=\frac{4!}{3!1!}=4\)

\(2*28*15*4 = 120*28 = 3360\)

Answer: D
_________________
Impossible is nothing to God.
Manager
Manager
avatar
Joined: 09 Apr 2012
Posts: 53
Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

Show Tags

New post 09 Feb 2013, 10:59
4
I always preferred slot method to solve these problems since childhood.

But my following approach gave me a wrong answer.
2 8 7 6 5 4 =13440.

What i forgot to do is divide (8*7) with 2 and 6*5 with another 2(since order doesnt matter between 2 defenders and 2 midfielders).

One of the other way of doing this is.

\(2P1\)\(*8P5\)/2*2.
Intern
Intern
avatar
Joined: 10 Dec 2013
Posts: 14
Location: India
Concentration: Technology, Strategy
Schools: ISB '16 (S)
GMAT 1: 710 Q48 V38
GPA: 3.9
WE: Consulting (Consulting)
Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

Show Tags

New post 28 Jan 2014, 11:40
Why cant we solve this by simple number theory?? If we use that approach then the answer changes
first we will have the 2 goalies, then 8 choices for first defender, then 7 choices for second defender and so on and so forth.
This would give us the answer as 2*8*7*6*5*4 = 13440

I know this is wrong answer but can someone please help me understand why this approach is wrong. Just because we are considering each position at a time?
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58464
Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

Show Tags

New post 29 Jan 2014, 08:45
Rohan_Kanungo wrote:
Why cant we solve this by simple number theory?? If we use that approach then the answer changes
first we will have the 2 goalies, then 8 choices for first defender, then 7 choices for second defender and so on and so forth.
This would give us the answer as 2*(8*7)*(6*5)*4 = 13440

I know this is wrong answer but can someone please help me understand why this approach is wrong. Just because we are considering each position at a time?


The point is that the number of ways to select 2 out of 8 is NOT 8*7=56 it's \(C^2_8=28\) and the number of ways to select 2 out of 6 is NOT 6*5=30 it's \(C^2_6=15\).

Hope it's clear.
_________________
Manager
Manager
avatar
Joined: 14 Nov 2011
Posts: 114
Location: United States
Concentration: General Management, Entrepreneurship
GPA: 3.61
WE: Consulting (Manufacturing)
GMAT ToolKit User
Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

Show Tags

New post 09 Jun 2014, 07:48
Bunuel wrote:
Rohan_Kanungo wrote:
Why cant we solve this by simple number theory?? If we use that approach then the answer changes
first we will have the 2 goalies, then 8 choices for first defender, then 7 choices for second defender and so on and so forth.
This would give us the answer as 2*(8*7)*(6*5)*4 = 13440

I know this is wrong answer but can someone please help me understand why this approach is wrong. Just because we are considering each position at a time?


The point is that the number of ways to select 2 out of 8 is NOT 8*7=56 it's \(C^2_8=28\) and the number of ways to select 2 out of 6 is NOT 6*5=30 it's \(C^2_6=15\).

Hope it's clear.



Hi Bunnel,

In this question why we not dividing by 3! ?

2c1 * {(8c2*6c2*4c1)/3!}

and why do we do so in the below question?

The Carson family will purchase three used cars. There are two models of cars available, Model A and Model B, each of which is available in four colors: blue, black, red, and green. How many different combinations of three cars can the Carsons select if all the cars are to be different colors?

a) 24
b) 32
c) 48
d) 60
e) 192
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58464
Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

Show Tags

New post 09 Jun 2014, 09:20
cumulonimbus wrote:
Bunuel wrote:
Rohan_Kanungo wrote:
Why cant we solve this by simple number theory?? If we use that approach then the answer changes
first we will have the 2 goalies, then 8 choices for first defender, then 7 choices for second defender and so on and so forth.
This would give us the answer as 2*(8*7)*(6*5)*4 = 13440

I know this is wrong answer but can someone please help me understand why this approach is wrong. Just because we are considering each position at a time?


The point is that the number of ways to select 2 out of 8 is NOT 8*7=56 it's \(C^2_8=28\) and the number of ways to select 2 out of 6 is NOT 6*5=30 it's \(C^2_6=15\).

Hope it's clear.



Hi Bunnel,

In this question why we not dividing by 3! ?

2c1 * {(8c2*6c2*4c1)/3!}

and why do we do so in the below question?

The Carson family will purchase three used cars. There are two models of cars available, Model A and Model B, each of which is available in four colors: blue, black, red, and green. How many different combinations of three cars can the Carsons select if all the cars are to be different colors?

a) 24
b) 32
c) 48
d) 60
e) 192


Because for this case the order of the selection matters: A = defense, B = midfield, C = forward is different from B = defense, A = midfield, C = forward, ... Therefore here we don't need factorial correction. While for "Carson family" problem Blue A/Black A/Red A is the same as Black A/Red A/Blue A...

Hope it's clear.

P.S. You can solve the second question with another approach described here: the-carson-family-will-purchase-three-used-cars-there-are-128876.html#p1056566
_________________
GMAT Club Bot
Re: Coach Miller is filling out the starting lineup for his indoor soccer   [#permalink] 09 Jun 2014, 09:20

Go to page    1   2    Next  [ 24 posts ] 

Display posts from previous: Sort by

Coach Miller is filling out the starting lineup for his indoor soccer

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne