Author 
Message 
TAGS:

Hide Tags

Current Student
Joined: 19 Nov 2012
Posts: 205
Concentration: Marketing, Social Entrepreneurship

Re: A Coach is filling out the starting lineup for his indoor [#permalink]
Show Tags
25 Oct 2013, 09:29
I arrived at D as well.
Quick question:
Obviously, the order is important here, which is why we do not account for repetitions.
How would the wording of the question change if order did not matter, which would lead us to divide D by n!?



Intern
Joined: 12 Dec 2012
Posts: 33
Concentration: Leadership, Social Entrepreneurship
GMAT 2: 660 Q48 V33 GMAT 3: 740 Q49 V41
GPA: 3.74

Re: Combination or Permutation: Can't make up my mind :) [#permalink]
Show Tags
06 Nov 2013, 23:00
Bunuel wrote: 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 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. Bunuel  I see how you arrived at D, but initially when I solved this problem I adjusted it because the order in which we make these selections shouldn't matter. Right? We are making a team of 6  so we need to select : 1GK (2C1), 2 Midfielders (8C2), 2 Defenders (6C2), 1 Forward (4C1) = 3360 (but aren't we over counting here, since it doesn't matter what order we make these selections in? So shouldn't we divide this by 4! to give us 140 different groupings? I made the mistake of NOT doing this on previous 'different grouping' questions and find it quite confusing. If you can explain when do adjust / when not to, it would be helpful! Cheers



Intern
Joined: 01 May 2013
Posts: 1

Re: A Coach is filling out the starting lineup for his indoor [#permalink]
Show Tags
16 Nov 2013, 07:34
You can alternatively come to the same conclusion using another approach:
First part  main team. There are 8 options (10 teammates excluding 2 goalkeepers). It means 8! combinations. But there are repeating elements. 2 defenders  that is 2!, 2 midfield  2!, and 3 will not be chosen and order inside of this unlucky team is also irrelevant, thereby 3!.
So we've got: 8! / (2!*2!*3!)
Second part is pretty easier  goalkeepers. There are two of them, one is to be chosen  2 options.
Now we have: 2*8! / (2!*2!*3!)



Intern
Joined: 10 Dec 2013
Posts: 20
Location: India
Concentration: Technology, Strategy
GPA: 3.9
WE: Consulting (Consulting)

Re: A Coach is filling out the starting lineup for his indoor [#permalink]
Show Tags
28 Jan 2014, 10:40
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
Joined: 02 Sep 2009
Posts: 43894

Re: A Coach is filling out the starting lineup for his indoor [#permalink]
Show Tags
29 Jan 2014, 07:45



Manager
Joined: 14 Nov 2011
Posts: 139
Location: United States
Concentration: General Management, Entrepreneurship
GPA: 3.61
WE: Consulting (Manufacturing)

Re: A Coach is filling out the starting lineup for his indoor [#permalink]
Show Tags
09 Jun 2014, 06:48
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
Joined: 02 Sep 2009
Posts: 43894

Re: A Coach is filling out the starting lineup for his indoor [#permalink]
Show Tags
09 Jun 2014, 08:20
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: thecarsonfamilywillpurchasethreeusedcarsthereare128876.html#p1056566
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  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



Manager
Joined: 30 Mar 2013
Posts: 127

Re: A Coach is filling out the starting lineup for his indoor [#permalink]
Show Tags
23 Sep 2014, 07:42
amkabdul wrote: Guys... since we have 6 positions in the field which are to be filled with 10 players.
Look at this: Filling the goal keeper position in 2 ways. Next: the rest of the 5 positions are to be filled with 8 players and each player at a different position would be 8p5 But since we have 2 positions each for the defence and mid. divide it with 2! twice.
So answer is : 2* (8P5/(2!*2!)) = 3360
A very simple and easy approach based on the basics of P 'n C
Kudo me if you like this. Agreed about the permutations approach. I was under the impression that we use permutations when there are places specified in the group, as there are here. I dont understand the combinations approach at all. For eg, in how many ways can you select a president and VP from a group of 6. Should be 6*5=30, or 6P2= 30. Can someone please point me in the direction where I can understand when to apply C and when to apply P. I understand that P is applied when order is imp, but the problem is, that when selecting president and VP from a group, order shouldn't be important. AB or BA is the same, and therefore, even those should be solved by using combinations. I hope I'm explaining my confusion well enough for someone to help me



GMAT Tutor
Joined: 24 Jun 2008
Posts: 1346

Re: A Coach is filling out the starting lineup for his indoor [#permalink]
Show Tags
23 Sep 2014, 09:38
usre123 wrote: the problem is, that when selecting president and VP from a group, order shouldn't be important. AB or BA is the same If you're choosing a President and VicePresident, and you choose A for President and B for VP, that's very different from choosing B for President and A for VP  you have a different President! So when you're choosing a President and VP, order is very important, and AB and BA are not the same selection.
_________________
GMAT Tutor in Toronto
If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com



Intern
Joined: 12 Nov 2014
Posts: 4

Re: A Coach is filling out the starting lineup for his indoor [#permalink]
Show Tags
21 Nov 2014, 02:21
Just wondering, why are we not just applying 2*8*7*6*5*4 ? Since the 8 remaining players can play anywhere, why do we need to differentiate Midfield vs defence vs forward?



NonHuman User
Joined: 09 Sep 2013
Posts: 13777

Re: A Coach is filling out the starting lineup for his indoor [#permalink]
Show Tags
29 Nov 2015, 03:53
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources



NonHuman User
Joined: 09 Sep 2013
Posts: 13777

Re: A Coach is filling out the starting lineup for his indoor [#permalink]
Show Tags
22 Jan 2017, 10:35
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources



Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2016

Re: A Coach is filling out the starting lineup for his indoor [#permalink]
Show Tags
25 Jan 2017, 10:22
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 We are given that there are 10 boys available to fill the following positions: 1 goalkeeper, 2 defenders, 2 midfielders, and 1 forward. We are also given that only 2 of the boys can play goalkeeper. Thus, we can select the goalkeeper in 2C1 = 2 ways. We now have 8 boys left and need to select 2 defenders from those 8 boys: 8C2 = (8 x 7)/2! = 28 ways We now have 6 boys left and need to select 2 midfielders: 6C2 = (6 x 5)/2! = 15 ways We now have 4 boys left and need to select 1 forward: 4C1 = 4 Thus, the number of ways to select 6 starters is: 2 x 28 x 15 x 4 = 3,360 Answer: D
_________________
Jeffery Miller
Head of GMAT Instruction
GMAT Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions



Intern
Joined: 06 Feb 2015
Posts: 2

Re: A Coach is filling out the starting lineup for his indoor [#permalink]
Show Tags
25 Feb 2017, 12:40
2C1 * 8C5 * 5C2 * 3C2 * 1C1
2C1  Goal keeper = 2 Remaining 5 people can be selected from 8C5 = 56
Now out of 5 people selected, we can select 2 defense, 2 mid field and 1 forward by 5C2 = 10 * 3C2 = 3 * 1C1 = 1
There fore 2* 56* 10*3*1 = 3360




Re: A Coach is filling out the starting lineup for his indoor
[#permalink]
25 Feb 2017, 12:40



Go to page
Previous
1 2
[ 34 posts ]



