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

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07 Nov 2013, 00:00

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]

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16 Nov 2013, 08:34

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]

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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?

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]

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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?

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...

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

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23 Sep 2014, 08:42

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

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

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21 Nov 2014, 03:21

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?

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

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29 Nov 2015, 04:53

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

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22 Jan 2017, 11:35

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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
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