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

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Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

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Updated on: 29 Apr 2019, 21:42
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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, 15:44.
Last edited by Bunuel on 29 Apr 2019, 21:42, edited 3 times in total.
Renamed the topic and edited the question.
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Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

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25 Oct 2009, 18:38
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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.
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Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

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25 Dec 2007, 16:48
4
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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.
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Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

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25 Dec 2007, 18:18
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
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Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

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05 Jan 2008, 20:04
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!
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Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

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05 Jan 2008, 23:27
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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
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Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

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25 Oct 2009, 17:08
5
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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

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

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25 Oct 2009, 17:57
2
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
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Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

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26 Oct 2009, 06:12
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
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Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

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02 Feb 2012, 20:30
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

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

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02 Feb 2012, 22:47
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.
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Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

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05 Feb 2012, 11:08
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
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05 Feb 2012, 11:31
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.
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Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

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05 Nov 2012, 20:07
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:
$$2 * \frac{8!}{2!6!} * \frac{6!}{2!4!} * \frac{4!}{1!3!} = 2 * \frac{8!}{2!2!3!}$$
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Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

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28 Dec 2012, 03:45
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? $$=\frac{4!}{3!1!}=4$$

$$2*28*15*4 = 120*28 = 3360$$

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

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09 Feb 2013, 09:59
4
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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.
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Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

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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: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

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29 Jan 2014, 07:45
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.
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Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

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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: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

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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: the-carson-family-will-purchase-three-used-cars-there-are-128876.html#p1056566
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