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Re: Team A and Team B are competing against each other in a game
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10 Jun 2018, 03:18
pkloeti wrote: Check the highlighted part of the stem:
Team A and Team B are competing against each other in a game of tugofwar. Team A, consisting of 3 males and 3 females, decides to lineup male, female, male, female, male, female. The lineup that Team A chooses will be one of how many different possible lineups? So does this mean in "GMAT language" that every 1st, 3rd and 5th person in the lineup MUST be a male? If so, I get it. I guess it's about practicing GMAT language...[/quote] Well, yes. How else? The stem directly specifies the desired lineup: it should start with a male and then alternate females and males.
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Re: Team A and Team B are competing against each other in a game
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10 Jun 2018, 03:24
Bunuel wrote: pkloeti wrote: Check the highlighted part of the stem:
Team A and Team B are competing against each other in a game of tugofwar. Team A, consisting of 3 males and 3 females, decides to lineup male, female, male, female, male, female. The lineup that Team A chooses will be one of how many different possible lineups? So does this mean in "GMAT language" that every 1st, 3rd and 5th person in the lineup MUST be a male? If so, I get it. I guess it's about practicing GMAT language... Well, yes. How else? The stem directly specifies the desired lineup: it should start with a male and then alternate females and males.[/quote] OK, thanks a lot!!



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Team A and Team B are competing against each other in a game
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09 Sep 2018, 10:31
Bunuel wrote: pkloeti wrote: Check the highlighted part of the stem:
Team A and Team B are competing against each other in a game of tugofwar. Team A, consisting of 3 males and 3 females, decides to lineup male, female, male, female, male, female. The lineup that Team A chooses will be one of how many different possible lineups? So does this mean in "GMAT language" that every 1st, 3rd and 5th person in the lineup MUST be a male? If so, I get it. I guess it's about practicing GMAT language... Well, yes. How else? The stem directly specifies the desired lineup: it should start with a male and then alternate females and males.[/quote] Hi Bunuel, Could you please clarify me the following? the I was also confused by the wording. But I was sure that it won't be 6! ways because MMMFFF can't be arranged in 6! ways. I thought it would be 6!/(3!*3!) = 20 But this option is not given. So, I had to figure out the solution by doing 3!*3! = 36 Now, I am not sure what the difference is between 6!/(3!*3!) = 20 and 3!*3! = 36. I think 6!/(3!*3!) = 20 means the number of the different possible combination. and 3!*3! = 36 means the arrangement with a certain condition. Please clarify the confusion.



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Team A and Team B are competing against each other in a game
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09 Sep 2018, 20:53
Raihanuddin wrote: Hi Bunuel,
Could you please clarify me the following? the
I was also confused by the wording. But I was sure that it won't be 6! ways because MMMFFF can't be arranged in 6! ways.
I thought it would be 6!/(3!*3!) = 20
But this option is not given. So, I had to figure out the solution by doing 3!*3! = 36
Now, I am not sure what the difference is between 6!/(3!*3!) = 20 and 3!*3! = 36.
I think 6!/(3!*3!) = 20 means the number of the different possible combination. and
3!*3! = 36 means the arrangement with a certain condition.
Please clarify the confusion. 3!*3! is the number of permutations when the lineup is male, female, male, female, male, female (MFMFMF) only. 6!/(3!3!) is the number of permutations of 3 males and 3 females without any restrictions.
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Re: Team A and Team B are competing against each other in a game
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09 Sep 2018, 21:06
Bunuel wrote: ashygoyal wrote: Team A and Team B are competing against each other in a game of tugofwar. Team A, consisting of 3 males and 3 females, decides to lineup male, female, male, female, male, female. The lineup that Team A chooses will be one of how many different possible lineups?
(A) 9 (B) 12 (C) 15 (D) 36 (E) 720
I have a doubt here. Should we count, (m1 w1 m2 w2 m3 w3) as a line up or, (m w m w m w) as a line up?
According to the question, it says one of the lineups is (m w m w m w) Doesn't that mean, other lineups will also include (m m m w w w), (w w w m m m),... and so on?
What wording in question makes us stick to only (m w m w m w)?
Please correct if my thought process is wrong.
Regards, Ashygoyal The question says that the lineup should be male, female, male, female, male, female (MFMFMF) only. So, alternating males and females, starting with a male. Any male can take any of the three M's positions and any female can take any of the three F's positions giving different arrangements. 3 men can be arranged in 3! ways and similarly 3 women can be arranged in 3! ways. So, the answer is 3!*3! = 36. Answer: D. Hope it's clear. Posted from my mobile deviceCan't we consider total combinations as 6!/(3!×3!) which gives 20. In this case with total combinations being 20, how can a specific combination gives 36 ?



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Re: Team A and Team B are competing against each other in a game
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09 Sep 2018, 21:10
anvesh004 wrote: Bunuel wrote: ashygoyal wrote: Team A and Team B are competing against each other in a game of tugofwar. Team A, consisting of 3 males and 3 females, decides to lineup male, female, male, female, male, female. The lineup that Team A chooses will be one of how many different possible lineups?
(A) 9 (B) 12 (C) 15 (D) 36 (E) 720
I have a doubt here. Should we count, (m1 w1 m2 w2 m3 w3) as a line up or, (m w m w m w) as a line up?
According to the question, it says one of the lineups is (m w m w m w) Doesn't that mean, other lineups will also include (m m m w w w), (w w w m m m),... and so on?
What wording in question makes us stick to only (m w m w m w)?
Please correct if my thought process is wrong.
Regards, Ashygoyal The question says that the lineup should be male, female, male, female, male, female (MFMFMF) only. So, alternating males and females, starting with a male. Any male can take any of the three M's positions and any female can take any of the three F's positions giving different arrangements. 3 men can be arranged in 3! ways and similarly 3 women can be arranged in 3! ways. So, the answer is 3!*3! = 36. Answer: D. Hope it's clear. Posted from my mobile deviceCan't we consider total combinations as 6!/(3!×3!) which gives 20. In this case with total combinations being 20, how can a specific combination gives 36 ? I think you did not read the whole discussion. Please do. I hope 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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Re: Team A and Team B are competing against each other in a game &nbs
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