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In how many ways can we seat 7 people in a round table with a certain 3 people not in consecutive order?

Guest Jan 21, 2016

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 #3
avatar+78557 
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Yeah, Melody....I see that I'm probably wrong.....I'm just trying to figure out my mistake....do you see it???   ....did I "double-count" somehow???

 

[ My answer seems  "logical"......LOL!!!! ]

 

 

 

cool cool cool

CPhill  Jan 21, 2016
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4+0 Answers

 #1
avatar+78557 
+5

Let's call the seats A,B,C,D,E,F,G

 

Then, the possibilities  for the chair positions for the three people are :

 

A C E       B D F       C E G

A C F       B D G

A D F       B E G

 

 

And each of the three people can be assigned to these seats in 3! ways = 6 ways

 

And the other 4 people can be arranged in 4! ways  = 24 ways for each arrangement above

 

So we have

 

7 position possibilities * 6 arrrangements of the three people for each position possibility* 24 arrangements of the other 4 people   = 1008 ways

 

As Melody often says....."That's what I think"

 

 

cool cool cool

CPhill  Jan 21, 2016
 #2
avatar+90970 
+5

In how many ways can we seat 7 people in a round table with a certain 3 people not in consecutive order?

 

I do not understand the question ://

What does it mean that the 3 people are not in consecutive order?

I will take it to mean that all 3 of these people cannot sit together.  I think CPhill has interpreted it the same way.

2 CAN sit together but not all 3

--------------------------------

ok, as CPhill said there are 3!=6 different ways these three can sit together - that is ignoring everyone else.

 

Mmm   Altogether there are 6! = 720 ways to sit 7 people at a table (rotations are counted as the same)  

So CPhill's answer can't be right.  Sorry Chris.   sad

 

My nexty task is to try and work out how many ways the group can be seated with the 3 siting next to each other.

Now if I tie the tree together and treat them like one person then there are 4 real people plus the triplet and these can be seated in 4! ways = 24 ways.

But there are 6 ways the triplets can be tired so that is  24*6=144 ways the three can be together

so

The number of ways that they can be seated if those three are not all together must be 720-144 = 576 ways

 

That is what I think anyway LOL

Melody  Jan 21, 2016
 #3
avatar+78557 
+5
Best Answer

Yeah, Melody....I see that I'm probably wrong.....I'm just trying to figure out my mistake....do you see it???   ....did I "double-count" somehow???

 

[ My answer seems  "logical"......LOL!!!! ]

 

 

 

cool cool cool

CPhill  Jan 21, 2016
 #4
avatar+90970 
+5

Chris, haven't you tried to put the three people together rather than NOT together ?

 

-----------------------

"7 position possibilities * 6 arrrangements of the three people for each position possibility* 24 arrangements of the other 4 people   = 1008 ways"

 

 

I am going to assume that is what you have tried to find. And I will look at your logic.

 

Yes there are 6 ways to seat the 3 that will remain together - in relation to each other.

So we will take one of these arrangements and seat these three people.

Noy you are right i believe, there are 4! = 24 arangements for the other 4

6*24=144     Yes so far I agree.

 

The you multiply this andwer by 7.....     Why did you do that ?
 

Melody  Jan 21, 2016

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