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In how many ways can you seat 40 students in 8 different benches (5 seats each), so that David and Arik won't sit on the same bench, and Larua will sit in the middle of the last bench?
This is what I think -

Case number 1 we have David sits on the last bench and Arik doesn't - $4\cdot35$
Case number 2 we have Arik sits on the last bench and David doesn't - $4\cdot35$
Case number 3 - neither Arik or David sit on the last bench - $35\cdot30$
Now we need to place the rest of the students - $37!$

So I get $2\cdot4\cdot35\cdot37! + 30\cdot35\cdot37! = 315\cdot37!$ ways overall. Did I miss anything?

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    $\begingroup$ It looks fine to me! $\endgroup$
    – String
    Dec 5, 2015 at 15:39

1 Answer 1

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David can sit in $40$ ways so arik can sit in $35$ ways laura has a fixed place. So $1$ way . Remaining can sit in $37!$ ways hence total ways are $40.35.1.37!=1400.37!$ . Hope its clear.

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  • $\begingroup$ I disagree... David can sit in 39 ways because Laura's seat is fixed. We can have different cases where David or Arik sit in the same bench with Laura. I don't see what's wrong with my answer $\endgroup$
    – Lisa
    Dec 5, 2015 at 15:38
  • $\begingroup$ I agree with @Lisa. $\endgroup$
    – String
    Dec 5, 2015 at 15:39
  • $\begingroup$ But its mentioned david and arik cNt be on same bench?? $\endgroup$
    – Archis Welankar
    Dec 5, 2015 at 15:44
  • $\begingroup$ won't sit on the same bench. $\endgroup$
    – true blue anil
    Dec 5, 2015 at 18:06

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