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Probability...a cracker

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by fightthegmat » Mon Sep 07, 2009 12:18 am
Two couples and one single person are seated at random in a row of five chairs.What is the probability that neither of the couples sits together in adjacent chairs?
A.1/5
B.1/4
C.3/8
D.2/5
E.1/2

how do you work on this? the tough part is 'couples to be seated NOT adjacent to each other'.
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by Nermal » Mon Sep 07, 2009 10:00 am
IMO B (1/4)

probability is the number of desired outcomes divided by the number of possible outcomes.

no. of possible outcomes: 5!=120
no. of desired outcomes: 5!/(2!*2!)=30
(I drew a line for each person and labelled the persons C1,C1,C2,C2,S)
30/120=1/4
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by rish » Mon Sep 07, 2009 10:09 am
IMO D

Probability for NO couple to sit together = 1 - (Probability of at least 1 couple sitting together)

Ways in which the first couple can sit together = 2*4! (1 couple is considered one unit,the couple sitting together can be arranged in 2 ways)
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 fightthegmat » Mon Sep 07, 2009 10:19 am
Nermal wrote:IMO B (1/4)

probability is the number of desired outcomes divided by the number of possible outcomes.

no. of possible outcomes: 5!=120
no. of desired outcomes: 5!/(2!*2!)=30
(I drew a line for each person and labelled the persons C1,C1,C2,C2,S)
30/120=1/4
The answer is D
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