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A Coach is filling out the starting lineup for his indoor

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Intern
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Re: A Coach is filling out the starting lineup for his indoor [#permalink] New post 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!?
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Re: Combination or Permutation: Can't make up my mind :) [#permalink] New post 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 :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.


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
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Re: A Coach is filling out the starting lineup for his indoor [#permalink] New post 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!)
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Re: A Coach is filling out the starting lineup for his indoor [#permalink] New post 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?
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Re: A Coach is filling out the starting lineup for his indoor [#permalink] New post 29 Jan 2014, 07:45
Expert's post
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.
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Re: A Coach is filling out the starting lineup for his indoor [#permalink] New post 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
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Re: A Coach is filling out the starting lineup for his indoor [#permalink] New post 09 Jun 2014, 08:20
Expert's post
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
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

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.


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Re: A Coach is filling out the starting lineup for his indoor   [#permalink] 09 Jun 2014, 08:20
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