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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 tug-of-war. 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 line-up: 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 tug-of-war. 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 line-up: 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 tug-of-war. 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 line-up: 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 (M-F-M-F-M-F) 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 tug-of-war. 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 (M-F-M-F-M-F) 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. Can'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 tug-of-war. 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 (M-F-M-F-M-F) 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. Can'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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Re: Team A and Team B are competing against each other in a game
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09 Sep 2018, 21:10
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