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by sanju09 » Sat Oct 22, 2011 3:46 am
Four men line up in a row. What is the probability that a certain two are next to each other?
(A) 1/6
(B) ¼
(C) 1/3
(D) ½
(E) 2/3
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by Anurag@Gurome » Sat Oct 22, 2011 3:59 am
sanju09 wrote:Four men line up in a row. What is the probability that a certain two are next to each other?
Total number of possible arrangements of 4 men = 4! = 24

Total number of possible combinations such that certain two are next to each other = 2*(3!) = 12

Hence, required probability = 12/24 = 1/2

The correct answer is D.
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by Anurag@Gurome » Sat Oct 22, 2011 3:59 am
sanju09 wrote:Four men line up in a row. What is the probability that a certain two are next to each other?
Total number of possible arrangements of 4 men = 4! = 24

Total number of possible combinations such that certain two are next to each other = 2*(3!) = 12

Hence, required probability = 12/24 = 1/2

The correct answer is D.
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by krishnakumar.ks » Sat Oct 22, 2011 4:04 am
Total number of ways of arranging these 4 persons is 4! = 24 ways.

Of these, if two people are always together, considering them as one unit, total number of arranging three people (one of them is a group of two who wanted to be together), is 3! * 2 = 12 ways.

Hence the required probability is 12/24 = 1/2.
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by vaibhavgupta » Sat Oct 22, 2011 4:56 am
sanju09 wrote:Four men line up in a row. What is the probability that a certain two are next to each other?
(A) 1/6
(B) ¼
(C) 1/3
(D) ½
(E) 2/3
u could also look at it like this :
X and y are two individuals.
scenario 1
X Y _ _ 2 ways
_ XY _ 2 ways
_ _ X Y 2 ways
Y X _ _ 2 ways
_ _ Y X 2 ways
_ Y X _ 2 ways

adding them up 12 ways
total ways 24.
hence probability = 1/2
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by zooki » Sat Oct 22, 2011 5:25 am
Say AB are the certain two among ABCD.
ABCD can be arranged 4!=24
[AB][C][D] can be arranged 3!=6 where AB can be arranged 2!=2

Therefore , the probability of AB being together is = (2*6)/24= 1/2
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