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Coach Miller is filling out the starting lineup for his
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Updated on: 20 Jun 2013, 01:37
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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 OPEN DISCUSSION OF THIS QUESTION IS HERE: acoachisfillingoutthestartinglineupforhisindoor85800.html
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Originally posted by jimjohn on 25 Dec 2007, 15:44.
Last edited by Bunuel on 20 Jun 2013, 01:37, edited 2 times in total.
Added the OA



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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 102 = 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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jimjohn wrote: thanks. that was from princeton but how do you know to break it up into: (8 C 2) * (6 C 2) * (4 C 1)
instead of just doing (8 C 5)
8c5 means there is one slot where we choose 5 from 8 elements.



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Re: PS permutations/combinations
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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
2C1 * 8C2*6C2*4C1
This is important to note, we have already chosen 2 goalkeeprs, so we are left with 8 people, and then after choosing 2 defense, we are left with 6 out of which to choose 2 midfield and then we are left with 4.
Hmm...but a good question... and it can be missed during the real exam! If proper attention is not paid!



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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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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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vscid wrote: jimjohn wrote: thanks. that was from princeton but how do you know to break it up into: (8 C 2) * (6 C 2) * (4 C 1)
instead of just doing (8 C 5) bmwhype2 I still have not understood why it can't be 8c5. Can you explain in detail?If you consider 8C5, you are missing all the possible positions of the players. In a group of 5 players the coach can built many teams just switching the positions of those 5 players. This problem has many approachs: 1st approach: n=2*C(8,5)*C(5,2)*C(3,2) = 3360; where you first consider two players for the goalkeeper position [2], second all the possible groups of 5 players from 8 players [C(8,5)] and third and last all the possible positions of all those 5 players [C(5,2)*C(3,2)]. 2nd approach: n=2*8!/3!= 13440, number to which you have to discount all the permutations among defenses and midfields, i.e. 2! and 2!. Therefore 13440/[2!·2!]=3360 3th approach: n=2*C(8,2)*C(6,2)*C(4,1)=3360; you just has to consider how many players you can fill the positions with. Since there is no difference between mildfield1 and mildfield2 you "count" combinations. If not, you should count permutations.



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Re: PS permutations/combinations
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25 Aug 2008, 12:27
i have small doubt ;
in the approach 2C18C26C24C1
why are selecting from 8 players whereas we will be left with 9 players after we made a selection for the goal keeper (as that is the logic stated in the solution for 6C2 i.e we are choosing from 6 because we will be left with 6 after selecting 2 from 8)



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Re: PS permutations/combinations
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13 Mar 2009, 12:13
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 2C1*8C5*5C2*3C2 = 3360 or 2C1*8C5*5C1*4C2 = 3360



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Re: PS permutations/combinations
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27 Sep 2009, 10:53
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
Soln: 1 goal keeper can be chosen from 2 boys in 2C1 ways. 2 defenders can be chosen from 8 boys in 8C2 ways 2 midfielders can be chosen from left over 6 boys in 6C2 ways 1 forwards can be chosen from the left over 4 boys in 4C1 ways
Thus total number of ways of choosing team is = 2C1 * 8C2 * 6C2 * 4C1 = 3360
Ans is D



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Re: PS permutations/combinations
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16 Feb 2010, 04:43
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 Goal Keeper selection = 2c1 Since only 2 can play at that position Defence Selection = 8c2 Midfield Selection = 6c2 Forward Selection = 4c1 Total combinations = 2c1 x 8c2 x 6c2 x 4c1 = 2 x 24 x 15 x 4 = 3360
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Re: Coach Miller is filling out the starting lineup for his
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04 Feb 2012, 23:50
I agree with walker  this is a permutation problem. As selecting 1 goalkeeper from 2 will result in 2 different teams. Likewise since each remaining players can play all positions  though order will matter, yet we need to divide by 2! for each positions to ensure there are no repetitions. Please correct me if I went wrong in my understanding.
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Re: Coach Miller is filling out the starting lineup for his
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05 Feb 2012, 00:27
sdas wrote: I agree with walker  this is a permutation problem. As selecting 1 goalkeeper from 2 will result in 2 different teams. Likewise since each remaining players can play all positions  though order will matter, yet we need to divide by 2! for each positions to ensure there are no repetitions. Please correct me if I went wrong in my understanding. I'm not sure I understand you point about the order. Anyway below is a different approach to this problem: 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 2C1 select 1 goalkeeper from 2 boys; 8C2 select 2 defense from 8 boys (as 2 boys can only play goalkeeper 102=8); 6C2 select 2 midfield from 6 boys (as 2 boys can only play goalkeeper and 2 we've already selected for defense 1022=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 1024=4) Total # of selection=2C1*8C2*6C2*4C1=3360 Answer: D. This problem is also discussed here: acoachisfillingoutthestartinglineupforhisindoor85800.htmlHope it helps.
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Re: Coach Miller is filling out the starting lineup for his
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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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Re: Coach Miller is filling out the starting lineup for his
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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
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18 Mar 2012, 22:13
Straight Answer: 2C1 * 8C2 * 6C2 * 4C1 But what if order is not considered and the coach needs 3 defense players.. Does anyone have answer.. Hope this question stimulates something thinking..
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Re: Coach Miller is filling out the starting lineup for his
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18 Mar 2012, 22:29
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, 3 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, without considering the order?
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Re: Coach Miller is filling out the starting lineup for his
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20 Jun 2013, 01:20
How many different groupings are possible? ...
This is a combination question based on the wording. The question asked for number of "Groups" and not the number of different plays.
Example, say the boys were b1, b2, b3, b4,...,b8 .. say you have group where b1 plays forward, b2 and b3 play mid, and b4, and b5 play back.. Now this is ONE group. This "group" does not change if only the player positions are changed.
Had the question been something like "Number of different plays", then I would go the permutation route which most people seem to agree with.
This is just my take, I maybe wrong in my assessment.
By my logic, the answer should be 112 [2 x 8C5] > we need to select 1 goalkeeper from 2 people and 5 folks from the remaining 8, per me the question does not care which positions each person plays in.
Thanks.



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Re: Coach Miller is filling out the starting lineup for his
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20 Jun 2013, 01:36
ashgoel wrote: How many different groupings are possible? ...
This is a combination question based on the wording. The question asked for number of "Groups" and not the number of different plays.
Example, say the boys were b1, b2, b3, b4,...,b8 .. say you have group where b1 plays forward, b2 and b3 play mid, and b4, and b5 play back.. Now this is ONE group. This "group" does not change if only the player positions are changed.
Had the question been something like "Number of different plays", then I would go the permutation route which most people seem to agree with.
This is just my take, I maybe wrong in my assessment.
By my logic, the answer should be 112 [2 x 8C5] > we need to select 1 goalkeeper from 2 people and 5 folks from the remaining 8, per me the question does not care which positions each person plays in.
Thanks. The correct answer is D. Your doubt is addressed here: acoachisfillingoutthestartinglineupforhisindoor85800.htmlHope it helps.
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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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