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Five people ( three women and two men ) arrange themselves in five consecutive seats. Find the probability that the men will be in two adjacent seats.

Please give me an answer as well as an explanation so I can learn from it. Thanks.

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Are there 2 men and 3 women? –  amWhy Jun 21 '13 at 21:39
    
@amWhy Yes sorry left that out –  MethodManX Jun 21 '13 at 21:40

3 Answers 3

up vote 2 down vote accepted

Assuming that each arrangement is equally probable.

Man A can sit in an end seat with $p=0.4$, in which case there is 1 seat for Man B which he can take with $p=0.25$, giving $p=0.1$ for this situation.

Man A can take a middle seat with $p=0.6$, giving Man B a $p=0.5$ chance of taking one of the two available seats, $p=0.3$.

Adding these together gives $p=0.4$.

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Consider "gluing" the two men together, so that there are really four slots to consider.

They can sit in seats 1&2, 2&3, 3&4, 4&5.

So we have essentially, four ways the two men (glued together) can sit next to each other, and then 3! ways in which the three women can take up the remaining seats 3. That gives us $4\cdot 3! = 4!$. In addition, the two man can swap positions, so we can "unglue" them, swap them, thus doubling the possibilities. This gives us $2\cdot 4!$ ways in which the men will be sitting together.

Now, there are $5!$ total possible combinations in which five total people can sit in 5 consecutive seats: 5 possible seats in which person one might sit, then 4 remaining seats where person two might sit, down to 3 seats options for person 3, etc until only one remaining seat where the fifth person must sit. This gives us $5\cdot 4\cdot 3 \cdot 2 \cdot 1 = 5!$ possible arrangements in which 5 people can sit.

So our probability $p$ that two men will sit next two each other will be the (number of arrangements in which two men must be seated together), divided by (the total number** of all possible seating arrangements, without restrictions), giving us: $$p = \dfrac {2\cdot 4!}{5!} = \dfrac 25 = 0.4$$

Of course, we are assuming here that every possible seating arrangement has equal probability: i.e., that the assignment of any one of the five people to any of five consecutive seats is random.

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there are 8 ways the men can be arranged not 4, you have neglected that Man A on the left is different to man B on the right –  Dale M Jun 21 '13 at 21:54
    
+1 for "gluing" the two men together. Hehe :) –  Prism Jun 21 '13 at 22:11
    
Thanks, @Prism ! –  amWhy Jun 21 '13 at 22:11
    
@amWhy: I was glued to this response! :-) +1 –  Amzoti Jun 22 '13 at 0:43

We will choose a pair of seats and put a "Reserved for Men" sign on it. The pair of seats can be chosen in $\binom{5}{2}$ equally likely ways.

Now think of the seats as X X X X X. There are $4$ ways to choose an adjacent pair of seats on which to put the Reserved signs. For the leftmost chosen seat can be any of the first four. Thus the required probability is $$\frac{4}{\binom{5}{2}}.$$

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