**What is the probability that 7 people can seat with a certain 3 people not in a consecutive order?**

Guest Nov 21, 2014

#3**+8 **

**What is the probability that 7 people can seat with a certain 3 people not in a consecutive order?**

Lets so those 3 people are glued together.

Then it would be like seating 4 people.

Sorry I can't count it would be 4 people and the cojoined triplets - that is 5 'bodies'. I will correct it from here.

there are 5! ways of sitting 5 people in a row = 120 ways

now thos 3 people can be joined together in any order so ther would be 3! ways = 6 ways.

So ther are 120*6= 720 ways those 3 people can be seated together.

now there are 7! ways of seating 7 people in a row. That is 5040 ways

5040-720=4320 so there are 4320 ways to seat everyone without those 3 individuals sitting together.

So the prob that they will not be sitting together is 4320/5040

$${\frac{{\mathtt{4\,320}}}{{\mathtt{5\,040}}}} = {\frac{{\mathtt{6}}}{{\mathtt{7}}}} = {\mathtt{0.857\: \!142\: \!857\: \!142\: \!857\: \!1}}$$

Now it is a running joke around here - none of us are really probablity experts. So don't trust me to much :))

CPhill answer and mine now agree. I believe this answer is correct

Melody Nov 21, 2014

#1**0 **

Are they sitting in a row or around a table?

Can 2 of those people sit together?

Melody Nov 21, 2014

#3**+8 **

Best Answer

**What is the probability that 7 people can seat with a certain 3 people not in a consecutive order?**

Lets so those 3 people are glued together.

Then it would be like seating 4 people.

Sorry I can't count it would be 4 people and the cojoined triplets - that is 5 'bodies'. I will correct it from here.

there are 5! ways of sitting 5 people in a row = 120 ways

now thos 3 people can be joined together in any order so ther would be 3! ways = 6 ways.

So ther are 120*6= 720 ways those 3 people can be seated together.

now there are 7! ways of seating 7 people in a row. That is 5040 ways

5040-720=4320 so there are 4320 ways to seat everyone without those 3 individuals sitting together.

So the prob that they will not be sitting together is 4320/5040

$${\frac{{\mathtt{4\,320}}}{{\mathtt{5\,040}}}} = {\frac{{\mathtt{6}}}{{\mathtt{7}}}} = {\mathtt{0.857\: \!142\: \!857\: \!142\: \!857\: \!1}}$$

Now it is a running joke around here - none of us are really probablity experts. So don't trust me to much :))

CPhill answer and mine now agree. I believe this answer is correct

Melody Nov 21, 2014

#5**0 **

Verymuch appreciated the effort :) im having problems in trying to analyze probability problems and btw thankyou again :) to follow up on a question the 3 guys cannot sit with each other right?

yuhki Nov 21, 2014

#6**0 **

if there are 4!*3!= 144 total ways those guys can sit why subtract it to 7! ? when only those 3! guys dont want to sit with other? im just not at this sorry for asking things like this?

yuhki Nov 21, 2014

#7**+3 **

Here's my contribution to this ....

The total arrangements for any 7 people sitting in a row is 7! = 5040

We can seat the 3 people together in five different ways.....(seats 1 - 5)

And for each of these there are 3! ways they can be arranged

So, the total number of ways they can sit together is:

(5 * 3!) = 30

So....the probability that they * don't* sit together is

1 - ( the probability they do sit together) =

1 - 30/5040 = 0.994047619047619 = about 99%

I don't know if this is correct......if anuone spots a flaw in my logic.....I'll be glad to know!!!

CPhill Nov 21, 2014

#8**0 **

You never need to appologise for admitting that you do not understand. we want you to learn. We do not to pretend to learn.

there are 144 ways those people can sit together. BUT the question asks you how many ways can they be seated so that they are NOT together. That is why I subtracted from the total number of seating arrangements possible which is 7!

Oh my way any 2 of those guys can sit together, just not all 3 of them. :)

Melody Nov 21, 2014

#9**0 **

i see :) how i think i kinda get it now. What if those 3 guys cannot sit with each other in any way, what would be the solution if each of the 3 dont want to be sitted to each other? what would you subtract to the total?

yuhki Nov 21, 2014

#10**+3 **

I had a post agreeing with you chris but it has disappeared and I have changed my mind.

I think I still like my answer. There are only 5 places those those 3 can sit, that is true, but you need to take to acount where everyone else is sitting as well. Anyway I am too tired to do more now. See you tomorrow. :))

Melody Nov 21, 2014

#11**+5 **

Mmmm...I see what Melody is saying.....I think that we * would* need to account for the other people, as well!!

For each arrangement of the three sitting together, there are 4! ways of arranging the other 4 people.

So, the total arrangements are 30*4! = 720

So

The probability of them not sitting together is

1 - 720/5040 = about 85.7%

I may think on this one some more !!!

CPhill Nov 21, 2014