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Tough Combinatoric Questions

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Tough Combinatoric Questions [#permalink]

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New post 14 Feb 2007, 21:08
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Can you tell the answer of the following questions?

1. A class of 30 students take a full row at a movie theater. There are six couples who must sit together. In how many different arrangements can the students sit?

2. A political debate has 8 pro-war attendees (PWAs) and 7 anti-war attendees (AWAs). The organizers want to seat the attendees next to each other, but with no two PWAs next to each other and no two AWAs next to each other. How many seating possibilities are there?

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New post 14 Feb 2007, 23:21
GMAT100 wrote:
Can you tell the answer of the following questions?

1. A class of 30 students take a full row at a movie theater. There are six couples who must sit together. In how many different arrangements can the students sit?

2. A political debate has 8 pro-war attendees (PWAs) and 7 anti-war attendees (AWAs). The organizers want to seat the attendees next to each other, but with no two PWAs next to each other and no two AWAs next to each other. How many seating possibilities are there?


1. tie six people and treat them like a 1 person. So, there are 24 + 1 people sitting at front row.

Possible arrangement = 25!
However, in a group of six, they can switch seats within the group
Total possible arrangement = 6! x 25!

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New post 15 Feb 2007, 06:57
devilmirror wrote:
[
1. tie six people and treat them like a 1 person. So, there are 24 + 1 people sitting at front row.

Possible arrangement = 25!
However, in a group of six, they can switch seats within the group
Total possible arrangement = 6! x 25!


just like to modify your answer to understand if i am getting the question right

6 couples so 12 people

30 -12+1 = 19

Possible arrangement = 19!
in a group of six, they can switch seats within the group = 6! * 25!
Again each couple can interchange thier place = 2! * 6! * 25!

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New post 15 Feb 2007, 08:36
I think you need to modify you solution. When they say six couples should sit together they mean 6 different couples. As in the requirement is only that each partner is paired with the other. hence 6 groups of 2

now : 18 singles, 6 couples (each couple can exchange seets)

= (18+6)! *2 should be the answer .. or am i missing some thing?

b) there is only one assignment PAPA...PAP but they can change with in a group = 8!7! (prowar!*anti!)

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New post 16 Feb 2007, 23:50
[quote="budugu"]I think you need to modify you solution. When they say six couples should sit together they mean 6 different couples. As in the requirement is only that each partner is paired with the other. hence 6 groups of 2

now : 18 singles, 6 couples (each couple can exchange seets)

= (18+6)! *2 should be the answer .. or am i missing some thing?

Should it be (18+6)! * 2^6 -- there are 6 couples

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New post 17 Feb 2007, 19:40
For #1, I agree with jainvineet's answer.

As for #2, the seating has to be PAPAPAPAPAPAPAP.
Ps can be arranged in 8! number of ways and As can be arranged in 7! number of ways.

Therefore the answer is 8! x 7!

Right?

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New post 21 Feb 2009, 09:30
Can someone tell OA to this question please.

A class of 30 students take a full row at a movie theater. There are six couples who must sit together. In how many different arrangements can the students sit?

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Re: Tough Combinatoric Questions [#permalink]

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New post 21 Feb 2009, 11:50
shmistry wrote:
Can someone tell OA to this question please.

A class of 30 students take a full row at a movie theater. There are six couples who must sit together. In how many different arrangements can the students sit?



Bundle each couple as one unit. So there are 6 units

30-12 = 18 singles + 6 couples = 24 units

24! ways

Each couple in 2ways

so 24! X 2!

6 such couples

24! X 2! X 6 is my answer

2nd one

I got 8 X 7 ^2 X 6 ^2 X .. 1^ 2, which is 8! X7!

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New post 21 Feb 2009, 11:59
why not
24! * (2! * 6) * (6! - for 6 positions among couples)

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Re: Tough Combinatoric Questions [#permalink]

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New post 21 Feb 2009, 13:10
MW MW MW MW MW MW S1 .. S18

treat each couple as one unit. there 24 units (6 coupls , 18 singles)

M and W can arrage in 2! for each couple.

Ans = 2!*2!*2!*2!*2!*2!*24!
= 2^6*24!
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New post 21 Feb 2009, 13:57
Thanks for the reply.

My question is why not multyple by another 6! for among couples.

24!*(2^6)*6!

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New post 22 Feb 2009, 14:32
shmistry wrote:
Thanks for the reply.

My question is why not multyple by another 6! for among couples.

24!*(2^6)*6!


You cannot because there are only 6 couples. factorial is applied as a whole not for n such couples

18 singles and 6 couples are considered as total and hence 24! ways. So the factorial you are looking for is taken care of

each couple can be arranged in 2! ways. if the question was about triplets and not couples it would be 3! ways. but still you would only multiply by 6 because there are only 6 of them

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Re: Tough Combinatoric Questions [#permalink]

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New post 07 Mar 2009, 00:03
icandy wrote:
each couple can be arranged in 2! ways. if the question was about triplets and not couples it would be 3! ways. but still you would only multiply by 6 because there are only 6 of them


jainvineet and x2suresh had the right answer above. You don't want to multiply by 6, you want to multiply by 2 six times; since you have 2 choices for how to arrange each of the six couples, you have 2x2x2x2x2x2 = 64 ways to arrange the six couples in total. That of course then needs to be multiplied by 24!.
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Re: Tough Combinatoric Questions   [#permalink] 07 Mar 2009, 00:03
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Tough Combinatoric Questions

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