BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

simple probability question

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |
User avatar
Legendary Member
Posts: 1132
Joined: Mon Jul 20, 2009 3:38 am
Location: India
Thanked: 64 times
Followed by:6 members
GMAT Score:760

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 :)



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

"Keep Walking" - Johnny Walker :P
Top
User avatar
Legendary Member
Posts: 1560
Joined: Tue Nov 17, 2009 2:38 am
Thanked: 137 times
Followed by:5 members

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
Top
User avatar
Legendary Member
Posts: 1560
Joined: Tue Nov 17, 2009 2:38 am
Thanked: 137 times
Followed by:5 members

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
Top
User avatar
Legendary Member
Posts: 1132
Joined: Mon Jul 20, 2009 3:38 am
Location: India
Thanked: 64 times
Followed by:6 members
GMAT Score:760

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
Top
Post new topic Post Reply
• Page 1 of 1