There are 5 couples. If they will sit 10 chairs in a row such that eac

Question

There are 5 couples. If they will sit 10 chairs in a row such that each couple sits side by side, how many possible cases are there?

A. 120 
B. 240 
C. 1,200 
D. 2,460 
E. 3,840

chetan2u · Accepted Answer

Hi,

Lets take each couple as a single identity, so total 5 places.. 
these 5 places can be arranged in 5! ways....
and each couple can be arranged in 2! within themselves..

so total ways =  5!*2!*2!*2!*2!*2! = 5!*2^5 = 120*32 = 3840 
E

BrentGMATPrepNow · Answer

Here's one approach: Take the task of seating the 10 people and break it into stages . Stage 1 : Select someone to sit in the 1st chair There are 10 people to choose from, so we can complete stage 1 in 10 ways Stage 2 : Select someone to sit in the 2nd chair The person seated in the 2nd chair must be the partner of the person in the 1st chair So we can complete this stage in 1 way. Stage 3 : Select someone to sit in the 3rd chair There are 8 people remaining. So, we can complete stage 3 in 8 ways Stage 4 : Select someone to sit in the 4th chair The person seated in the 4th chair must be the partner of the person in the 3rd chair So, we can complete this stage in 1 way. Stage 5 : Select someone to sit in the 5th chair There are 6 people remaining. So, we can complete stage 5 in 6 ways Stage 6 : Select someone to sit in the 6th chair The person seated in the 6th chair must be the partner of the person in the 5th chair So, we can complete this stage in 1 way. Stage 7 : Select someone to sit in the 7th chair There are 4 people remaining. So, we can complete stage 7 in 4 ways Stage 8 : Select someone to sit in the 8th chair The person seated in the 8th chair must be the partner of the person in the 7th chair So, we can complete this stage in 1 way. Stage 9 : Select someone to sit in the 9th chair There are 2 people remaining. So, we can complete stage 9 in 2 ways Stage 10 : Select someone to sit in the 10th chair One 1 person remains. So, we can complete this stage in 1 way. By the Fundamental Counting Principle (FCP), we can complete all 10 stages (and thus seat all 10 people) in (10)(1)(8)(1)(6)(1)(4)(1)(2)(1) ways ( = 3840 ways ) Answer: E -------------------------- 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-counting/video/775 You can also watch a demonstration of the FCP in action: https://www.gmatprepnow.com/module/gmat ... /video/776 Cheers, Brent

adiagr · Answer

Tie the couples. Then we have 5 persons that have to be arranged in 5 places. 5! ways.

Now the couples can change position with each other. 2! ways.

5! * (2!)^2 = 120 *32 = 3840

Answer is E.

5! * 2! = 240 ways.

MathRevolution · Answer

5!(2^5)=3,840. Hence, the correct answer is E.

fantaisie · Answer

Could anyone perhaps explain me the logic behind the highlighted part? I went on and just multiplied 5! by 2 and still cannot understand why it would be  2^5 ... Thank you!

FacelessMan · Answer

Each couple can be arranged in 2 ways within themselves. And there are 5 couples. So a total of 2*2*2*2*2 = 2^5 combinations.

sashaman14 · Answer

Best way to explain this is that normally we would just arrange 5 groups/people in 5 sections as (5 * 4 * 3 * 2 * 1), grouping the couples up since they sit next two each other.
But in this case, each section/group of chairs/ couple / etc can be arranged two different ways (A1/A2 or A2/A1), so it 
ends up being (10 * 8 * 6 * 4 * 2)

MathRevolution · Answer

Since the couples can exchange their seats, we mupltiply by 2. Then, since there are 5 couples, we multiply 2 five times. Hence, 2^5.

hsbinfy · Answer

perfect !

solved this way only

BrushMyQuant · Answer

Bunuel : Please untag probability

Bunuel · Answer

______________
Done. Thank you.

djangobackend · Answer

total 10 seats, 
say M are men and F are female
so M1 has 10 options and consecutively F1 has only 1 option since M1F1 should sit together 
then M2 has 8 options and so on. 
total 10*8*6*4*2 = 3840 ways

OR 
C1, C2, C3, C4, C5 couples can pick seats in 5! ways and each couple can arrange themselves in 2! ways so total ways = 5! * (2!)^5 = 3840 ways

There are 5 couples. If they will sit 10 chairs in a row such that eac

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There are 5 couples. If they will sit 10 chairs in a row such that eac

Prep Toolkit

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