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?
The answer is 2/5. I couldn't find a quick beautiful solution.
Let's numerate people 1,2,3,4,5 where (1,2) are a couple, (3,4) are a couple, 5 is a single person.
there is 1/5 probability that the single person is going to sit in chair one, then we need to put 1,2,3,4 in such a way that 2 people from the same couple don't sit next to each other. If we choose any person to sit in the second chair, his partner has to sit in chair 4 - probability of that is 1/3. Probability 1/5*1/3 = 1/15
there is 4/5 probability that a person from 1-4 is going to sit in chair one. Let's say it's person 1. then it's either person 5 who is going to sit in chair 2 (probability of that 1/4, then person 2 has to sit in chair 4 - probability of that 1/3 - total probability 4/5*1/4*1/3 = 1/15) or either 3 or 4 [say 3](probability of that 2/4 = 1/2, then his partner can't sit in chair 3 - probability of that 2/3 - total probability 4/5*1/2*2/3 = 4/15)
1/15+1/15+4/15 = 6/15=2/5. Not a beautiful solution but it took me less than 1.5 min to solve it.
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walker, can you explain me this formula in details? thanks
P^4_4 - the number of permutations of the first couple and 3 single persons
P^2_2 - the number of permutations of 2 single persons in the first couple
P^4_4*P^2_2 - the total number of permutations in which the first couple sits together.
P^4_4*P^2_2 - the total number of permutations in which the second couple sits together.
P^3_3*P^2_2*P^2_2 - the total number of permutations in which the first couple sits together and the second couple sits together.
P^4_4*P^2_2+P^4_4*P^2_2-P^3_3*P^2_2*P^2_2 - the total number of permutations in which the first couple sits together or the second couple sits together.
P^5_5 - the total number of permutations of 5 single persons
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I also did by first finding the probability of couples being together and then subtracting from 1.
2 couples together and 1 single [A1A2][B1B2][C] = 3!*2*2 = 24 ways 3! for ABC and 2 each for interchanging positions between each couple
1 couple and 3 singles (subtract ways that have been counted in above) = [A1A2]BCD = [4! * 2] - 24 = 24 ways (this needs to be multiplied by 2 as we have two couples). So, 24*2=48
a bit I just return in quant to relax after hard (very hard!) work in Verbal....
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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?
Choices are 1/5, 1/4 , 3/8, 2/5 and 1/2
Total possible ways = 5! Two couple sits together [A1A2][B1B2]C = 2! * 2!*3! =24 only first couple sit together [A1A2] B1 B2 C = 2!*4! - 24 (substract 24 two couple sits together) =24 only second couple sit together A1A2 [B1 B2] C =2!*4! - 24 =24
probability = 1 - 24*3/120 = 1-3/5 =2/5
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Total possible ways of selecting 5 persons in different ways is = 5!
Two couple sits together [A1A2][B1B2]C. So it forms only 3 groups. we can arrange the groups in 3! ways. we have 2 ways to arrange the person in a couple. so we have 2!=2 and so the calculation is Two couple sits together [A1A2][B1B2]C = 2! * 2!*3! =24
only first couple sit together [A1A2] B1 B2 C. so it forms 4 groups (the couple and the other persons). we can arrange the different groups in 4! ways. we have 2 ways to arrange the person in a couple. so we have 2!=2 = 2!*4! - 24 (substract 24 two couple sits together) =24 only second couple sit together A1A2 [B1 B2] C, so it forms 4 groups. we can arrange the different groups in 4! ways. =2!*4! - 24 =24
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?
Choices are 1/5, 1/4 , 3/8, 2/5 and 1/2
Soln: Lets take the positions to be _ _ _ _ _
Now considering that the single person sits in seat 1. The remaining seats can be filled in = 4 * 2 * 1 * 1 = 8 ways
Considering single person in seat 2 The remaining seats can be filled in = 4 * 2 * 1 * 1 = 8 ways
Considering single person in seat 3 The remaining seats can be filled in = 4 * 2 * 2 * 1 = 16 ways
Considering single person in seat 4 The remaining seats can be filled in = 4 * 2 * 1 * 1 = 8 ways
Considering single person in seat 5 The remaining seats can be filled in = 4 * 2 * 1 * 1 = 8 ways
Total number of ways in which 5 people can be arranged is 5! ways
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?
Choices are 1/5, 1/4 , 3/8, 2/5 and 1/2
Probability both couples sit together = 3! x 2! x 2! = 24
Probability that only first couple sits together = 4! x 2! - both couple sits together = 48 - 24 = 24 Probability that only second couple sits together = 4! x 2! - both couple sits together = 48 - 24 = 24
Therefore no couple sits together =1 - (24 x 3) / 5! = 1 - 3/5 = 2/5
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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?
Choices are 1/5, 1/4 , 3/8, 2/5 and 1/2
There are 5! = 120 ways to seat the 5 people.
I'll use the same naming of the people. (1,2) (3,4) and 5 as the single.
5 _ _ _ _ = _ _ _ _ 5 Now how many ways can we sit people without couples touching?
4*2*1*1 = 8
and since 5 _ _ _ _ = _ _ _ _ 5 we have 8*2 = 16 ways to seat people without couples touching when the single is on the end.
_ 5 _ _ _ = _ _ _ 5 _ Now how many ways can we sit people without couples touching?
4*2*1*1 = 8
and since we have two seating arranges for that to work 8*2 = 16 ways to seat people when the single is sitting one spot from the end.
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?
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?
walker, can you explain me this formula in details? thanks
P^4_4 - the number of permutations of the first couple and 3 single persons
P^2_2 - the number of permutations of 2 single persons in the first couple
P^4_4*P^2_2 - the total number of permutations in which the first couple sits together.
P^4_4*P^2_2 - the total number of permutations in which the second couple sits together.
P^3_3*P^2_2*P^2_2 - the total number of permutations in which the first couple sits together and the second couple sits together.
P^4_4*P^2_2+P^4_4*P^2_2-P^3_3*P^2_2*P^2_2 - the total number of permutations in which the first couple sits together or the second couple sits together.
P^5_5 - the total number of permutations of 5 single persons
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Math: GMAT Math Book ||| General: GMATTimer ||| Chicago Booth: Slide Presentation The People Who Are Crazy Enough to Think They Can Change the World, Are the Ones Who Do.
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