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Five math coaches A, B, C, D, E are seated in a row at a banquet.  How many possible seatings are there, if the coaches A and D cannot be seated next to each other.

 Dec 20, 2019
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Note that the total possible arrangements are  5!  =  120

 

Let's  look at the arrangements where AD do sit together

AD - - -

-AD - -

- - AD -

- - - AD

 

So  they could occupy  4 different positions....and for each of these they could be seated in 2 ways (AD or DA)

 And for each of these.....the other 3 can be seated in 3! ways

 

So.....the number of seatingss where they don't  sit together  =

 

120  - (4)(2)(3!)  =

 

120   - 8*6  =

 

120  - 48  =

 

72

 

cool cool cool

 Dec 21, 2019

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