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Two couples and one single person are seated at random in a [#permalink]

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02 Apr 2006, 16:30

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Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?

Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?

Thanks

I got 9/10
let's assume we have 5 people numbered 1 to 5 and we want to keep

(1,2) and (4,5) together therefore 3 must be first, in the middle or last

and we can order (1,2) in two ways
(4,5) in two ways

the single person has to sit either first middle or last

case 1

first

1(because there is only 1 single person )*4(because any of the other 4 people can sit next to him ) * 2 (because out of the three only 2 that are not coupled to 4 can sit next to 4) *1( because only the person who isnt coupled to 2 can sit next to 2) and times 1 because only one person is left ) = 8

case 2
last

4*2*1*1*1 =8

case 3
and middle

4*2*1*2*1= 16

now total possible arrangements is 5! = 120

(16+8+8)/120 = 32/120 and this is = to 4/15

hope i didnt screw it up !

and 9/10 just seems a bit too close to 1 to be correct