3 couples...Round Table...PLEASE HELP ME!

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landlover

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Hi. I'm sorry to inconvenience you, but could somebody please help me with this math problem? It would be greatly appreciated! ;)

Okay, so the problem isn't exactly the math, but figuring out how work out the problem. What are the steps I need to take? We didn't exactly get any lesson on how to do it, we pretty much just got handed the problem and told to do it. No one in my entire class has figured out the answer yet.

The Problem is:

Three couples (6 people) are randomly seated at a round table. what is the probability that the members of at least one of the couples are seated next to each other?

Thanks :D
 
landlover said:
Hi. I'm sorry to inconvenience you, but could somebody please help me with this math problem? It would be greatly appreciated! ;)

Okay, so the problem isn't exactly the math, but figuring out how work out the problem. What are the steps I need to take? We didn't exactly get any lesson on how to do it, we pretty much just got handed the problem and told to do it. No one in my entire class has figured out the answer yet.

The Problem is:

Three couples (6 people) are randomly seated at a round table. what is the probability that the members of at least one of the couples are seated next to each other?

Thanks :D
Click to expand...

You must have done problem involving permutations/combinations - correct?

What is the total number of ways you can sit sixpeople around a table?

If the textbook or teacher does not help - do a google search!

Please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.
 
so theres 120 ways of arranging them, because theres six people so you would multiply 5*4*3*2*1=120.

Thats where im stuck. How do i figure out how many of those arrangements involve at least one couple sitting together?
 
landlover said:
so theres 120 ways of arranging them, because theres six people so you would multiply 5*4*3*2*1=120.

Thats where im stuck. How do i figure out how many of those arrangements involve at least one couple sitting together?
Click to expand...

Now think of it this way...

One couple will be tied together - and rest (4) may sit randomly.....
 
I'm sorry im making everything so complicated but i really am just clueless on this :(
 
Would the answer be .15? I multiplied 3*2*1 on all 3 couples because it would mean all the possible combinations for them to sit. so that got me 18, and then i divided that by 100.....so is that right? :?:
 
Since the problem asks for 'at least one' couple sitting next to one another, find the probability that NO couple sits next to each other and subtract from 1.

Three couples are seated at a round table.

Sit one wife down anywhere ( 6 choices). Her husband can sit opposite her or two seats away (3 out of 5 choices).

Sit the second wife down in any empty seat(1 out of 4 choices). Her husband has two choices out of three.

The last couple can sit anywhere, as long as it's not together. Only one choice out of three will let the final couple sit without being together.

So, \(\displaystyle \frac{1}{6}\cdot \frac{3}{5}+\frac{1}{4}\cdot \frac{2}{3}=\frac{4}{15}\) is the probability none of the couples sit together.

So, \(\displaystyle \frac{15}{15}-\frac{4}{15}=\frac{11}{15}\) is the probability that at least one couple sits together.

You can also think of it as \(\displaystyle \frac{2^{5}}{5!}\).
 
Thank you so much! I finally was able to understand what someone was talking about! Thanks alot and God bless you! :D
 
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