+0  
 
0
56
3
avatar

In how many ways can 8 people be seated in a row of chairs if two of the people, John and Paul, refuse to sit next to each other?

 Dec 22, 2020
 #1
avatar
+1

Well I'm really tired but here we go, let us use complementary counting, so that is 

$total$ $outcomes$ - unfavorable$ $outcomes$, total outcomes is all ways that can be seated without restrictions so that is

8!= $40320$, then the unfavorable outcomes is if John and Paul sit next to each other, we can pretend that the pair is one person and that is how many ways to arrange 7 people, which is 7!, that results in $5040$

 

Subtracting, we get:

 

$40320-5040$ = $35280$
 

Please check, as I possibly have missed somethings but that's the idea.

 Dec 22, 2020
 #2
avatar
+1

  I have no idea if the above answer is correct.....hope it is

     here is my take on it

          there are   8! seating arrangements = 40320

             there are 14 ways those two knuckleheads could be seated next to each other ( you can reverse their order to get 14)

                 those 14 ways leave 6! for the rest of the folks to be seated

40320 - 14 * 6! = 30240 ways  

    I do not know if this is correct.....it MIGHT be...it might NOT be !  

 Dec 22, 2020
 #3
avatar+114320 
+1

Let's suppose  that  they can sit together

 

John and Paul can occupy  any of 7 positions and for each of these,they can sit in 2 ways...and for all these  arrangements  the other people can  sit  in 6!  ways

So....the total arrangements where they sit together =  7 * 2 * 6!  = 10080 ways

 

And the total possible arrangements  = 8!   = 40320

 

So....the ways  which they don't sit together  =   total arrangements - those where they sit  together  =   40320  - 10080   = 30240

 

 

cool cool cool

 Dec 22, 2020

25 Online Users

avatar
avatar