Team A and Team B are competing against each other in a game

Question

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

(A) 9
(B) 12
(C) 15
(D) 36
(E) 720

cbh · Accepted Answer

The question is really confusing. It's not clear if the suggested approach to line up 3 males and 3 females is a pattern or just one of the possible solutions. 
I also resolved it through 6! as the questions asked " The  lineup that Team A chooses  will be o ne of how many different  possible lineups"

Even if you google the question you will find a google book, and the response is still confusing:)) It says Any of the 3 males can be first in the line and any of the 3 females can be second
But why not the females start the line? I cannot find this constraint in the question

Bunuel · Answer

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.

Answer: D.

Hope it's clear.

ScottTargetTestPrep · Answer

We need to determine the number of ways to lineup male, female, male, female, male, female.

Since there are 3 males, we have 3 options for the first spot, and since there are 3 females, we have 3 options for the second spot. Then we have 2 options for the third spot, 2 options for the fourth, and 1 option for each of the last two spots. Thus, the number of ways to lineup that group is 3 x 3 x 2 x 2 x 1 x 1 = 36.

Answer: D

pushpitkc · Answer

There is only one way of forming the line MFMFMF.
However, there are 3! ways of arranging the men in the line
If there are 3 men (M1,M2,M3) 
the ways they can be arranged are:
1st 2nd 3rd
M1-M2- M3
M1- M3- M2
M2- M1- M3
M2- M3- M1
M3- M1- M2
M3- M2- M1(Total 6 ways)

Similarly women can also be arranged in 6 ways.

Total possibilities of arranging both men and women are 6*6*1 = 36 ways(Option D)

amanvermagmat · Answer

Men have to occupy 1st, 3rd and 5th positions. This can be done in 3*2*1 = 6 ways
Women have to occupy 2nd, 4th and 6th positions. This can be done in 3*2*1 = 6 ways

Together, the above two things have to be done which can be done in 6*6 = 36 ways

Hence  D answer

Luckisnoexcuse · Answer

Please correct me if i am wrong

M1, M2, M3, F1, F2, F3 can be arranged in 6! ways

Sent from my iPhone using GMAT Club Forum

amanvermagmat · Answer

Hi

when you say 6!, you are listing ALL the possible arrangements of 3 men and 3 women in a straight line. 
But as per the question, we have constraints. Men can occupy only 1st, 3rd, 5th places from start while women can occupy only 2nd, 4th, 6th places from start.

BrentGMATPrepNow · Answer

Take the task of lining up the 6 competitors and break it into   stages  .

Stage 1 : Select a competitor for the 1st position 
This person must be a male. 
Since there are 3 males to choose from, we can complete stage 1 in  3  ways

Stage 2 : Select a competitor for the 2nd position 
This person must be a female. 
Since there are 3 females to choose from, we can complete stage 2 in  3  ways

Stage 3 : Select a competitor for the 3rd position 
This person must be a male. 
There are 2 males remaining to choose from (since we already selected a male in stage 1), so we can complete stage 3 in  2  ways

Stage 4 : Select a competitor for the 4th position 
This person must be a female. 
There are 2 females remaining to choose from. So we can complete stage 4 in  2  ways

Stage 5 : Select a male for the 5th position 
There's only 1 male remaining. So we can complete stage 5 in  1  way

Stage 6 : Select a female for the 6th position 
There's only 1 female remaining. So we can complete stage 6 in  1  way

By the Fundamental Counting Principle (FCP), we can complete all 6 stages (and thus create a 6-person lineup) in  (3)(3)(2)(2)(1)(1)  ways (= 36 ways)

Answer:  D

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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ashygoyal · Answer

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

cbh · Answer

Sorry Bunuel, I would disagree, the question doesn't say  should be made  and  only  it clearly says that's why so many people found it's confusing.
I get it, GMAT candidates should learn the GMAT language otherwise risk of failure on such questions is high.

Bunuel · Answer

In this context "decides to lineup" = " should lineup".

pkloeti · Answer

Hi  Bunuel ,
where does it state in the question that the lineups MUST start with a male? It is just stated that Team A chooses M-F-M-F-M-F. If this is one of all possible lineups, Team A could also start with a female (F-M-F-M-F-M therefore IS a possible lineup). What I mean is "the option" to start with F is there, it's not discarded in the question stem.

What am I not seeing?

Thanks!

Bunuel · Answer

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?

(A) 9
(B) 12
(C) 15
(D) 36
(E) 720

pkloeti · Answer

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?

(A) 9
(B) 12
(C) 15
(D) 36
(E) 720

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...

Bunuel · Answer

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.

Raihanuddin · Answer

Well, yes. How else? The stem directly specifies the desired line-up: it should start with a male and then alternate females and males.

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.

Bunuel · Answer

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.

Michele4 · Answer

Bunuel   chetan2u  why can't we use the classic formula for arrangements here "6!/3!*3!" ??

chetan2u · Answer

\frac{6!}{3!3!}  is the formula to arrange 6 things out of which 3 are of one kind and the other 3 are of the other kind.
SO this will include mmffmf or mmmfff etc. But we are looking at mfmfmf only. here m have specific places and f have specific places. They can be arranged within these places, so 3!*3!

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