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Two couples and one single person are seated at random in a [#permalink]

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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?

let x - a single person, 0 - a person of couples. 1 - a person of first couples, 1 - a person of second couples

x0000 16 cases to seat together at least one couple. (x1122 - 4,x2112 - 4,x2211 - 4, 1221 - 4)
0x000 16 cases to seat together at least one couple.
00x00 8 cases to seat together at least one couple.
000x0 and 0000x 16*2 due to symmetry.

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

Please show the steps

P for both couples sit together (C1C1)(C2C2)(S) = 2!2!3! =24 P for first couple sit together and other couples not (C1C1)(C2C2S) = 2!4! - 2!2!3! =24 similarly for P second couple sit together and others couples not =24

let x - a single person, 0 - a person of couples. 1 - a person of first couples, 1 - a person of second couples

x0000 16 cases to seat together at least one couple. (x1122 - 4,x2112 - 4,x2211 - 4, 1221 - 4) 0x000 16 cases to seat together at least one couple. 00x00 8 cases to seat together at least one couple. 000x0 and 0000x 16*2 due to symmetry.

Therefore,

p=1-(16*4+8)/5!=1-8*9/5*4*3*2=1-3/5=2/5

But I've wasted a lot of time to find the way.

are there any shortcuts?

Here my take.. seems short to me.

Consider first couple as a unit and so sits together. The # of arrangements 2*4! (2 ways for couple, 4! for the couple as a unit and the remaining 3 ppl)

Consider second couple as a unit and so sits together. The # of arrangements 2*4! (2 ways for couple, 4! for the couple as a unit and the remaining 3 ppl).

Consider each couple as a unit. The # of arrangements 2*2*3! (2 ways for each couple and 3! for couples a units and the loner)

So total number of ways in which atleast one couple sits together is 2*4! + 2*4! - 2*2*3! (AUB = A+B-AB)

Total number of possible ways of seating 5!

Prob that no couple sits togehter = 1 - (4*4! - 4*3!)/5! = 2/5

let x - a single person, 0 - a person of couples. 1 - a person of first couples, 1 - a person of second couples

x0000 16 cases to seat together at least one couple. (x1122 - 4,x2112 - 4,x2211 - 4, 1221 - 4) 0x000 16 cases to seat together at least one couple. 00x00 8 cases to seat together at least one couple. 000x0 and 0000x 16*2 due to symmetry.

Therefore,

p=1-(16*4+8)/5!=1-8*9/5*4*3*2=1-3/5=2/5

But I've wasted a lot of time to find the way.

are there any shortcuts?

Here my take.. seems short to me.

Consider first couple as a unit and so sits together. The # of arrangements 2*4! (2 ways for couple, 4! for the couple as a unit and the remaining 3 ppl)

Consider second couple as a unit and so sits together. The # of arrangements 2*4! (2 ways for couple, 4! for the couple as a unit and the remaining 3 ppl).

Consider each couple as a unit. The # of arrangements 2*2*3! (2 ways for each couple and 3! for couples a units and the loner)

So total number of ways in which atleast one couple sits together is 2*4! + 2*4! - 2*2*3! (AUB = A+B-AB)

Total number of possible ways of seating 5!

Prob that no couple sits togehter = 1 - (4*4! - 4*3!)/5! = 2/5

I took <1m too..
_________________

Your attitude determines your altitude Smiling wins more friends than frowning

let x - a single person, 0 - a person of couples. 1 - a person of first couples, 1 - a person of second couples

x0000 16 cases to seat together at least one couple. (x1122 - 4,x2112 - 4,x2211 - 4, 1221 - 4) 0x000 16 cases to seat together at least one couple. 00x00 8 cases to seat together at least one couple. 000x0 and 0000x 16*2 due to symmetry.

Therefore,

p=1-(16*4+8)/5!=1-8*9/5*4*3*2=1-3/5=2/5

But I've wasted a lot of time to find the way.

are there any shortcuts?

Here my take.. seems short to me.

Consider first couple as a unit and so sits together. The # of arrangements 2*4! (2 ways for couple, 4! for the couple as a unit and the remaining 3 ppl)

Consider second couple as a unit and so sits together. The # of arrangements 2*4! (2 ways for couple, 4! for the couple as a unit and the remaining 3 ppl).

Consider each couple as a unit. The # of arrangements 2*2*3! (2 ways for each couple and 3! for couples a units and the loner)

So total number of ways in which atleast one couple sits together is 2*4! + 2*4! - 2*2*3! (AUB = A+B-AB)

Total number of possible ways of seating 5!

Prob that no couple sits togehter = 1 - (4*4! - 4*3!)/5! = 2/5

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

Soln: Probability that none will sit together is = (8 + 8 + 16 + 8 + 8)/120 = 48/120 = 2/5

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