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19 Aug 2006, 14:53
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Let's worry about the single fellow/gal S at the end. Imagine he arrives late and ensure that the couples are seperate! Again, think of two cases: ( abbrev. C1 man or wife of one couple, C2 man or wife of other couple)
(1) C1 C2 C1 C2...........4*2*1*1=8 ways
(2) C1 C2 C2 C1..........8 ways
(Notice that C1 C1 C2 C2 is unworkable)
Where could/must S sit in each case? For (1), in any of 5places. For (2) only in one place. So total=8*5+8=48
Required Probability= 48/5! 4*3*4/5*4*3*2= 2/5
(1) C1 C2 C1 C2...........4*2*1*1=8 ways
(2) C1 C2 C2 C1..........8 ways
(Notice that C1 C1 C2 C2 is unworkable)
Where could/must S sit in each case? For (1), in any of 5places. For (2) only in one place. So total=8*5+8=48
Required Probability= 48/5! 4*3*4/5*4*3*2= 2/5
20 Aug 2006, 00:05
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kevincan
at teh third position...
but why have u ignored the following cases :
(3) C2 C1 C1 C2
(4) C2 C1 C1 C2
.. also.. can we not think of this problem has a "letter arrangement problem" with 3 alphabets... (C1C1)(C2C2)(thrird person)..these can be arranged in 3! ways..
Further, each of the 2 couples can be individually arranged in 2! ways..
meaning 2! for (C1male , C1female)
meaning 2! for (C2male , C2female)
.. so we have = 3! x 2! x 2!
Now, the prob. of couples siitting 2gether is = 3! x 2! x 2!/5! = 1/5
So, teh prob. of couples not sitting 2getehr is = 1- 1/5=4/5
.. however, teh answer is 2/5 ...
can you please guide me ... is my approach wrong..
