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Difficult Math Problem #15

This topic has 3 member replies

Forum: Difficult Math Problem #15

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OA coming after some people attempt an answer:

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?

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Senior | Next Rank: 100 Posts Default Avatar
Joined
03 Sep 2006
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Total possibilites: 5!

if i arrange couple A with single B like this:

AAB

then i can put couple C in the places marked by '_' :

_A_A_B_

so ill have 4P2 = 12 options (A is man and female i assume Laughing so thats why it is 4P2)

i can make 6 like this, so i have 72 options

so total there are 72 options to arrange couples to not sit by each other out of 5!

72/120 = 3/5

i hope im right Razz

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let AA BB C stand for 5 people
if C sit on the first chair, 4 choices for the second chair and 2 choice for the third chair = 4*2= 8.
Same thing happens if C sit on the fifth chair = 8
Now, if C sit on the second chair, 4 choices (for the first chair) * 2 (for the third) = 8. Similarly, if C sit on the fourth chair, 4*2 = 8
finally, if C in the middle (third chair) 4 choices for the first*2 choices for the second*2 choices for the fourth = 16
Total = 8+8+8+8+16 = 48
Probability = 48/5! = 2/5 = 40%
Hope right

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Master | Next Rank: 500 Posts Default Avatar
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here's the OA:

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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