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

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25 Oct 2009, 16:43

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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 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)

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.
_________________

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

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25 Oct 2009, 18:55

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

For that matter we could have gone an extra step and said it should be 9c6 (leaving other goalie out) , but for Bunuel explanation ) Bunuel is awesome.. of course this time I was clear too and I in fact prepared a similar example as Bunuel
_________________

Thanks, Sri ------------------------------- keep uppp...ing the tempo...

Press +1 Kudos, if you think my post gave u a tiny tip

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

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05 Feb 2011, 06:40

i made the same mistake as rvthryet... thanks for the explanation even tho i still dont understand how it makes such a different. i can c from your example that it happens. but i cannot understand the logic behind it. at the end he is choosing 5 ppl out of 8. the order have no effect.

but from what u are saying - unless its very clear that i need to choose only 5 ppl from a group without ANY distinction - it will be 5C8...

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

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12 Nov 2011, 02:17

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

Answer: D.

Sorry for bringing up an old post. I am clear with the above explanation except for one fact . Is there a necessity to choose in the above order. What happens if I choose the goalie, then the forward and then the defense & midfield. In such a case the combinations change drastically.

2C1 * 8C1 * 6C2 *4C2 = 1440.

The questions does not specify any order with selecting one group over another. Can someone please explain this?

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

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12 Nov 2011, 15:32

gmatrant wrote:

Sorry for bringing up an old post. I am clear with the above explanation except for one fact . Is there a necessity to choose in the above order. What happens if I choose the goalie, then the forward and then the defense & midfield. In such a case the combinations change drastically.

2C1 * 8C1 * 6C2 *4C2 = 1440.

The questions does not specify any order with selecting one group over another. Can someone please explain this?

That would still work. Except that there is a mistake in the counting you did 2C1 * 8C1 * 7C2* 5C2 = 3360

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

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04 Jan 2012, 22:00

Nice question. Easy to make the 8C5 trap. The individual selections need to be made and that, IMO, is the key to this problem. Answer is D.
_________________

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

Answer: D.

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

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.
_________________

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

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05 Nov 2012, 20:07

2

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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: \(2 * \frac{8!}{2!6!} * \frac{6!}{2!4!} * \frac{4!}{1!3!} = 2 * \frac{8!}{2!2!3!}\)

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

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17 Nov 2012, 10:52

I did it in the next way: 1) ways to chose Goalkeepers - it's obvious 1C2=2 2) ways to chose Field players within the group - comes from DDMMF p=5!/2!2!=30 3) ways to chose Field players from 8 possible options - 5C8=56

Multiple everything: 2*30*56=3360

However the previous comment is the most elegant
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Re: Combination or Permutation: Can't make up my mind :) [#permalink]

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24 Dec 2012, 01:06

144144 wrote:

i made the same mistake as rvthryet... thanks for the explanation even tho i still dont understand how it makes such a different. i can c from your example that it happens. but i cannot understand the logic behind it. at the end he is choosing 5 ppl out of 8. the order have no effect.

but from what u are saying - unless its very clear that i need to choose only 5 ppl from a group without ANY distinction - it will be 5C8...

thanks bunuel.

With the 8C5 logic, you are not accommodating the case where 2 defenders are different from 2 midfielders.

To pick up on the 4 player example Bunuel gave, 4 players - A,B,C,D we should choose 1 for defense and 1 for forward. (no restrictions).

When we do 4C2 Defence-forward AB AC AD BC BD CD

But we are missing the case CA where C is defender and A is forward. In this particular example we can use the nPr permutation formula.

Back to original question, Bunuels method is pretty kickass.
_________________

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

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28 Dec 2012, 03:45

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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

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? \(=\frac{4!}{3!1!}=4\)

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

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25 Oct 2013, 02:20

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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.

gmatclubot

Re: A Coach is filling out the starting lineup for his indoor
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25 Oct 2013, 02:20

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