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simple probability question

Expert replies
Source: — Problem Solving |

by harsh.champ » Fri Feb 19, 2010 3:43 am
daretodream wrote:TWO couples and a single person are to be seated on 5 chairs such that no
couple is seated next to each other. What is the probability of the above??
Let the couples be A1,A2 and B1,B2.

Now,let one A ,one B and the single person be S be seated - 3! x 2C1 x 2C1.

So,the other two A and B can be seated in 2 x 2 ways.

Hence, the answer would be 96 ways.
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



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by thephoenix » Fri Feb 19, 2010 3:46 am
daretodream wrote:TWO couples and a single person are to be seated on 5 chairs such that no
couple is seated next to each other. What is the probability of the above??
Ways in which the first couple can sit together = 2*4! (1 couple is considered one unit)
Ways for second couple = 2*4!
These cases include an extra case of both couples sitting together
Ways in which both couple are seated together = 2*2*3! = 4! (2 couples considered as 2
units- so each couple can be arrange between themselves in 2 ways and the 3 units in 3!
Ways)
Thus total ways in which at least one couple is seated together = 2*4! + 2*4! - 4! = 3*4!
Total ways to arrange the 5 ppl = 5!
Thus, prob of at least one couple seated together = 3*4! / 5! = 3/5
Thus prob of none seated together = 1 - 3/5 = 2/5
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by thephoenix » Fri Feb 19, 2010 3:48 am
harsh.champ wrote:
daretodream wrote:TWO couples and a single person are to be seated on 5 chairs such that no
couple is seated next to each other. What is the probability of the above??
Let the couples be A1,A2 and B1,B2.

Now,let one A ,one B and the single person be S be seated - 3! x 2C1 x 2C1.

So,the other two A and B can be seated in 2 x 2 ways.

Hence, the answer would be 96 ways.
the q is abt probabilty
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by harsh.champ » Fri Feb 19, 2010 4:47 am
thephoenix wrote:
harsh.champ wrote:
daretodream wrote:TWO couples and a single person are to be seated on 5 chairs such that no
couple is seated next to each other. What is the probability of the above??
Let the couples be A1,A2 and B1,B2.

Now,let one A ,one B and the single person be S be seated - 3! x 2C1 x 2C1.

So,the other two A and B can be seated in 2 x 2 ways.

Hence, the answer would be 96 ways.
the q is abt probabilty
OIC.
I forgot to consider the sample space.
[If the option choices had been given,I could have rectified my mistake then and there only-This is also one of the imp. reasons why option choices should be given alongwith the question]
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



Just because something is hard doesn't mean you shouldn't try,it means you should just try harder.

"Keep Walking" - Johnny Walker :P
Join the discussion