GMAT OG Team A and Team B are competing against

AbeNeedsAnswers 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

D

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. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat- ... /video/775

You can also watch a demonstration of the FCP in action: https://www.gmatprepnow.com/module/gmat ... /video/776

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html


MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/permutation-t122873.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/combinations-t123249.html


Cheers,
Brent

Hi AbeNeedsAnswers,

We're told that Team A consists of 3 males and 3 females and is going to lineup male, female, male, female, male, female. We're asked for the total number of different possible lineups that Team A can create.

Since we're putting elements 'in order', we're dealing with a permutation....
(M)(F)(M)(F)(M)(F)

For the first spot, there are 3 options.
For the second spot, there are 3 options.
For the third spot, since we've already placed one of the men, there are only 2 options.
For the fourth spot, since we've already placed one of the women, there are only 2 options.
For the fifth spot, there's only 1 option left.
For the sixth spot, there's only 1 option left.
Now we just have to multiply all of those numbers together...

(3)(3)(2)(2)(1)(1) = 36 possible options

Final Answer: D

GMAT assassins aren't born, they're made,
Rich