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quick permutations question (1 Viewer)

ianc

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hi all

I have done a question, but haven't got the solution, so was wondering could someone give me an answer?

There are 8 candidates to take an exam, 3 in Maths and 5 in English. In how many ways can they be seated in a row if the 3 Maths candidates cannot sit together?

Thanks!
 
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pLuvia

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Ways of the maths candidates sitting together=6!x3!
Ways of the maths candidates not sitting together=8!-(6!x3!)=36000

Edit: Mistake, thanks Riviet
 
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Riviet

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pLuvia - you forgot that the 3 maths guys can be rearranged in 3! ways.

.'. no. of ways = 8! - (6! x 3!)
 

haque

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if dealing with the case where all three are separated then the above two answers are wrong, however the way the question is stated it would imply that 2 people would be allowed together.?
 

Riviet

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haque said:
if dealing with the case where all three are separated then the above two answers are wrong, however the way the question is stated it would imply that 2 people would be allowed together.?
Hmm good point, I think we would assume that 2 of them adjacent would be counted. If not, then I guess we would subtract those arrangements.
 

haque

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based on the way its worded ur answer should be right riviet
 

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