In order to continue enjoying our site, we ask that you confirm your identity as a human. Thank you very much for your cooperation.
Answer
Hint: Here, first we have to take the couples as A, B and C. First we have to calculate the number of possible ways that the couples are seated next to each other. Here, each couple has to be arranged among themselves in 2! ways. Each couple can be seated next to each other in 3! ways. With this data we have to calculate the total possible seating arrangements. Next, to find the number of ways that all ladies sit together, we have to arrange the ladies among themselves which is 3! ways and then, consider 3 men and 3 ladies as 4 units where 3 ladies are tied together which will be 4! Ways.Complete Step-by-Step solution:Here, we are given that the married couples are to be seated next to each other in a cinema hall.Now, we have to find the number of possible ways they can be seated and also the number of ways of their seating if all ladies sit together.First let us consider the three couples as A, B and C which each couple is taken as single unit as in figure:
For $A_2$ you have a good idea, but you missed a detail. There are $3$ ways to choose which couple does not sit together; this is the factor that you missed. Then there are $3$ ways to split that couple (your $P_1aP_2b$, $aP_1bP_2$, and $aP_1P_2b$), $2$ ways to decide which couple is $P_1$ and which is $P_2$, and $2$ ways to seat the members of each couple, so
$$A_2=3\cdot3\cdot2\cdot2^3=9\cdot16=144\;.$$
In terms of your calculation, that changes $4!-2!\cdot3$ to $4!-3\cdot2!\cdot3=6$, and $2!^3\cdot3\cdot6=144$.
Probably the easiest way to calculate $A_3$ is to subtract the total of the other three possibilities from $6!$, since you already have them.
Added: Since there are only three couples, $A_4$ isn’t too hard to calculate by hand. Let $A$ be the person in the first seat in the row; there are $6$ choices for $A$. Let $A'$ be $A$’s spouse. Anyone except $A'$ can sit in the second seat, so there are $4$ possibilities; call the person who sits there $B$. Now split the count into two cases.
The third seat is taken by $C$, a member of the third couple. There are $2$ ways to choose $C$, so there are altogether $6\cdot4\cdot2$ ways to fill the first three seats with members of three distinct couples. The fourth seat can be occupied by $A'$ or $B\,'$, and in either case the last two people can sit in either order, so there are $2^2=4$ ways to fill out the row, for a total of $6\cdot4\cdot2\cdot4=192$ arrangements.
The third seat is taken by $A'$, so that we have $ABA'$ in the first three seats. The rest of this calculation is spoiler-protected to give you a chance to finish it on your own.
$A'$’s spouse is in the first seat, so at first sight it might appear that any of the $3$ remaining people can take the fourth seat. However, if $B\,'$ takes it, the third couple will be forced to sit together, so it must actually be taken by a member of the third couple, say $C$, and the last three seats must be filled $CB\,'C'$: the only choice is in the order of the two members of the third couple. This case therefore accounts for another $6\cdot4\cdot2=48$ arrangements. Combining results, we see that $A_4=192+48=240$.
Manager
Joined: 14 Dec 2004
Posts: 72
Three married couples have bought 6 seats in a row for a [#permalink]
Three married couples have bought 6 seats in a row for a performance of a musical comedy.1. In how many ways can they be seated?2. In how many ways can they be seated if each couple is to sit together with the husband to the left of the wife?3. In how many ways can they be seated if each couple is to sit together?
4. In how many ways can they be seated if all the men are to sit together and all the women are to sit together?
Manager
Joined: 22 Jun 2004
Posts: 231
Location: Bangalore, India
Re: Three married couples have bought 6 seats in a row for a [#permalink]
Here are my answers. (1) 6 people in 6 seats => 6! = 720 (2)There is a definite order contsraint that husband is to the left of wife. Thus, 3 couples can seat in 3! = 6 ways. (3)3 couples can seat in 3! ways and each couple can themselves change seats in 2! ways. Thus, the totall number of ways = 3! * 2! *2! * 2! = 48 (4)3 Men can seat 3! ways in 3 seats. 3 Women can seat 3! ways in 3 seats. Men and women as two packs can occupy the seats in 2! ways (Leftmost 3 seats women or rightmost 3 seats women). => the total number of ways = 2! * 3! * 3! = 72 Awaiting answers..
cloaked_vessel wrote:
Three married couples have bought 6 seats in a row for a performance of a musical comedy.1. In how many ways can they be seated?2. In how many ways can they be seated if each couple is to sit together with the husband to the left of the wife?3. In how many ways can they be seated if each couple is to sit together?
4. In how many ways can they be seated if all the men are to sit together and all the women are to sit together?
Senior Manager
Joined: 07 Jun 2004
Posts: 461
Location: PA
Re: Three married couples have bought 6 seats in a row for a [#permalink]
1. 6! 2. 3! 3. 3! * (2!)^3
4. 3! * 3! * 2!
Manager
Joined: 14 Dec 2004
Posts: 72
Re: Three married couples have bought 6 seats in a row for a [#permalink]
correct answers:720,6,48,
72
Non-Human User
Joined: 09 Sep 2013
Posts: 25390
Re: Three married couples have bought 6 seats in a row for a [#permalink]
Hello from the GMAT Club BumpBot!Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________
Re: Three married couples have bought 6 seats in a row for a [#permalink]
11 Dec 2018, 02:47
Moderators:
Math Expert
10656 posts
Math Expert
87884 posts