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3 persons - one couple and one single are seated at random in a row of 5 seats

what is the probability that the couple does not sit together

please explain your approach

best regards

Let's find the probability that a couple sits together (right next to each other) and subtract that value from 1.

Total # of ways 3 persons \(C_1\), \(C_2\) and \(S\) to be seated in a row of 5 seats is \(\frac{5!}{2!}=60\). Consider this, we are interested in arrangement of \(C_1, \ C_2, \ S, \ E, \ E\), so in arrangement of 5 letters out of which 2 E's are identical (E denotes an empty seat);

# of ways for a couple to sit together is \(\frac{4!}{2!}*2=24\). Consider a couple as a single unit: \(\{C_1,C_2\}, \ S, \ E, \ E\), so we have total of 4 units out of which 2 E's are identical, # of arrangement of these units is \(\frac{4!}{2!}\), but \(C_1\), \(C_2\) within their unit can be arranged in 2 ways (\(\{C_1,C_2\}\) or \(\{C_2,C_1\}\)), so total # of arrangement for this case is \(\frac{4!}{2!}*2=24\);

Re: 3 persons (1 couple and 1 single) are seated at random in a [#permalink]

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28 Dec 2012, 18:42

keiraria wrote:

3 persons (1 couple and 1 single) are seated at random in a row of 5 chairs. What is the probability that the couple does not sit together?

A. 5/7 B. 4/5 C. 2/5 D. 3/5 E. 11/18

Given: {H} {W} {S} will attempt to seat on _ _ _ _ _ seats

How many ways for {H} {W} {S} to seat on _ _ _ _ _ seats? 5*4*3 How many ways for {HW} {S} to seat together on _ _ _ _ _ seats? 4*3 Then multiply by 2 to account for the arrangement of HW. 4*3*2!

What is the probability of {H}{W} NOT seating together? \(=1 - \frac{4*3*2}{5*4*3} = 1 - \frac{2}{5} = 3/5\)

The single person is no different from an empty chair

Thus, there are \(\frac{{5*4}}{2}=10\) ways to pick two chairs for the couple, but only 4 in which they sit together (CCEEE, ECCEE, EECCE, EEECC).

1 - 4/10 = 3/5 is the answer.

The perspective we often use to solve such questions is this: The vacant seats are no different from two identical people. Assume that each vacant spot is taken by an imaginary person V.

'The single person is no different from an empty chair' is a refreshing perspective that we can use! Good point! Assume the rest of the three chairs are vacant. Since it is a probability question, the probability we will obtain will be correct. Mind you, the number of ways in which you can arrange a couple and an individual is not given by 10. It is given by 60 only (as shown by Bunuel above). But 10 is the number of ways in which we can choose 2 seats for a couple. In 4 of those 10 ways, the seats will be next to each other and in 6 cases they will not be. Hence the probability obtained will be 3/5. You can also think that you can make the husband and the wife sit in 2 of the 5 chairs in 5*4 = 20 ways. Out of these, in 4*2 = 8 ways, they will sit next to each other. In 12 ways, they will not sit next to each other. So probability will still remain 12/20 = 3/5

As discussed before, in probability questions, whatever logic you use to get the numerator, use the same logic to get the denominator.
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3 persons (1 couple and 1 single) are seated at random in a [#permalink]

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23 Oct 2015, 21:15

why for this question are we doing 5!/2! for the chairs to people?

I recall from a circular question, when there was 4 men and 4 women sitting around the table, and we had to calculate the number of ways if all the women sat together.

which was (4-1)! * 4C1 * 4!

So why is this question 5!/2! instead of 5C3 ?

We can incorporate this question into my confusion surrounding this "http://gmatclub.com/forum/there-are-x-people-and-y-chairs-in-a-room-where-x-and-y-are-170525.html"

There are x people and y chairs in a room where x and y are positive prime numbers. How many ways can the x people be seated in the y chairs (assuming that each chair can seat exactly one person)?

(1) x + y = 12. Since x and y are primes, then x=5 and y=7 OR x=7 and y=5.

If x=5 and y=7, the number of arrangements would be C57∗5!=7!2!, where C57 is the number of way to choose 5 chairs from 7, and 5! is the number of arrangements of 5 people on those chairs.

If x=7 and y=5, the number of arrangements would be C57∗5!=7!2!, where C57 is the number of way to choose 5 people who will get the chairs, and 5! is the number of arrangements of 5 people on those chairs.

Re: 3 persons (1 couple and 1 single) are seated at random in a [#permalink]

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25 Oct 2015, 15:14

1

This post received KUDOS

GMATDemiGod wrote:

why for this question are we doing 5!/2! for the chairs to people?

I recall from a circular question, when there was 4 men and 4 women sitting around the table, and we had to calculate the number of ways if all the women sat together.

which was (4-1)! * 4C1 * 4!

So why is this question 5!/2! instead of 5C3 ?

We can incorporate this question into my confusion surrounding this "http://gmatclub.com/forum/there-are-x-people-and-y-chairs-in-a-room-where-x-and-y-are-170525.html"

There are x people and y chairs in a room where x and y are positive prime numbers. How many ways can the x people be seated in the y chairs (assuming that each chair can seat exactly one person)?

(1) x + y = 12. Since x and y are primes, then x=5 and y=7 OR x=7 and y=5.

If x=5 and y=7, the number of arrangements would be C57∗5!=7!2!, where C57 is the number of way to choose 5 chairs from 7, and 5! is the number of arrangements of 5 people on those chairs.

If x=7 and y=5, the number of arrangements would be C57∗5!=7!2!, where C57 is the number of way to choose 5 people who will get the chairs, and 5! is the number of arrangements of 5 people on those chairs.

I think you are confusing between straight up arrangements of n things in a row = n! and arrangement of n people around a table = circular arrangement = (n-1)!

For the question : 3 people (1 couple and 1 single), we are doing 5!/2! to account for 5 'things' to arrange (C1,C2, S,E,E where Es are empty seats) and 2! accounts for the 2 empty seats.

You can not do 5C3 as this will remove the cases when you have EEC1C2S or EC1C2ES etc.

Re: 3 persons (1 couple and 1 single) are seated at random in a [#permalink]

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28 Dec 2017, 05:36

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