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.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: Combination or Permutation: Can't make up my mind :) [#permalink]
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!

Re: A Coach is filling out the starting lineup for his indoor [#permalink]
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.

Re: A Coach is filling out the starting lineup for his indoor [#permalink]
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?

Re: A Coach is filling out the starting lineup for his indoor [#permalink]
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.

Re: A Coach is filling out the starting lineup for his indoor [#permalink]
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?

Re: A Coach is filling out the starting lineup for his indoor [#permalink]
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...