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6 persons are going to theater and will sit next to each oth

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6 persons are going to theater and will sit next to each oth  [#permalink]

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New post 16 Jun 2010, 20:08
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6 persons are going to theater and will sit next to each other in 6 adjacent seats. But Martia and Jan can not sit next to each other. In how many arrangement can this be done

I understood that the restriction must be deal first
by finding the number of way the restriction happen and remove from the total number of way to arrange the n !


It is 2 ! for arrangement and 4 ! for the remaining 4 people

but what I don t understand is why it is time by 5 as the OA gives


I saw some other type like that

For instance digit 1,2,3,4,5 IF EACH DIGIT is used only once how many ways can each digit be arranged such 2 and 4 are not adjacent -
In this case the restriction is 2!x4! not multiplied by anything else
Can anyone explain me why


thanks for your Time


regards
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Re: Permutations with Restriction  [#permalink]

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New post 16 Jun 2010, 20:45
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Total number of arrangements = 6!

Assuming the two sit next to each other, we have 5!x2 arrangements (This is because when they are sitting next to each other, we can consider both of them as one "unit" and hence there are 5 units, i.e them and the other 4 people. This will lead to an arrangement of 5! and then between them, they can be seated in two ways, so it's 2x5!)

So answer = Total - Arrangements with them sitting next to each other

= 6! - 2x5! = 5! x 4 = 480

I think you assumed they were sitting next to each other and did only the 2! and 4!

What are the numbers given in the official answer? I believe you only mentioned word choices.
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Re: Permutations with Restriction  [#permalink]

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New post 16 Jun 2010, 20:54
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Maude wrote:
Hello

Please need help to understand the following

6 persons are going to theater and will sit next to each other in 6 adjacent seats
But Martia and Jan can not sit next to each other .In how many Arrangement can this be done

I understood that the restriction must be deal first
by finding the number of way the restriction happen and remove from the total number of way to arrange the n !


It is 2 ! for arrangement and 4 ! for the remaining 4 people

but what I don t understand is why it is time by 5 as the OA gives


I saw some other type like that

For instance digit 1,2,3,4,5 IF EACH DIGIT is used only once how many ways can each digit be arranged such 2 and 4 are not adjacent -
In this case the restriction is 2!x4! not multiplied by anything else
Can anyone explain me why


thanks for your Time


regards


Hi, and welcome to Gmat Club! Below is the solution for your problem.

You are right saying that probably the best way to deal with the questions like this is to count total # of arrangements and then subtract # of arrangements for which opposite of restriction occur. But the way you are calculating the later is not correct.

Total # of arrangements of 6 people (let's say A, B, C, D, E, F) is \(6!\).
# of arrangement for which 2 particular persons (let's say A and B) are adjacent can be calculated as follows: consider these two persons as one unit like {AB}. We would have total 5 units: {AB}{C}{D}{E}{F} - # of arrangement of them 5!, # of arrangements of A and B within their unit is 2!, hence total # of arrangement when A and B are adjacent is \(5!*2!\).

# of arrangement when A and B are not adjacent is \(6!-5!*2!\).

In your example about 5 digits the answer would be:
Total # of arrangements of 5 distinct digits is \(5!\).
# of arrangement for which 2 digits 2 and 4 are adjacent is: consider these two digits as one unit like {24}. We would have total 4 units: {24}{1}{3}{5} - # of arrangement of them 4!, # of arrangements of 2 and 4 within their unit is 2!, hence total # of arrangement when 2 and 4 are adjacent is \(4!*2!\).

# of arrangement when 2 and 4 are not adjacent is \(5!-4!*2!\).

Hope it helps.
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Re: 6 persons are going to theater and will sit next to each oth  [#permalink]

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New post 22 Dec 2015, 02:12
Maude wrote:
6 persons are going to theater and will sit next to each other in 6 adjacent seats. But Martia and Jan can not sit next to each other. In how many arrangement can this be done


6 person can sit on 6 seats in way = 6*5*4*3*21 = 6! = 720

6 person can sit on 6 seats in ways such that Martia and Jan sit next to each other = (5*4*3*2*1)*(2!) = 5!*2! = 240

Favorable cases = 720 - 240 = 480
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Re: 6 persons are going to theater and will sit next to each oth  [#permalink]

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New post 22 Dec 2015, 03:40
the quick formular for this kind of questions is solve removing the constraint - the opposite of the constraint.
Hence, 6! - 5!
the opposite of the constraint is if the two guys must sit next to each other, that is, they are one, hence you have 5.
how many ways can you arrange 6 unique letters minus how many ways can you arrange 5 unique letters
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Re: 6 persons are going to theater and will sit next to each oth  [#permalink]

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New post 22 Dec 2015, 04:06
Nez wrote:
the quick formular for this kind of questions is solve removing the constraint - the opposite of the constraint.
Hence, 6! - 5!
the opposite of the constraint is if the two guys must sit next to each other, that is, they are one, hence you have 5.
how many ways can you arrange 6 unique letters minus how many ways can you arrange 5 unique letters


Hi,
you have found the correct method but missed out the final step..
both are taken as one and we get the arrangement as 5! but within the two both can be arranged in 2! ways..
so answer becomes 6!-5!*2
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Re: 6 persons are going to theater and will sit next to each oth  [#permalink]

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New post 07 Sep 2018, 11:01
Maude wrote:
6 persons are going to theater and will sit next to each other in 6 adjacent seats. But Martia and Jan can not sit next to each other. In how many arrangement can this be done

I understood that the restriction must be deal first
by finding the number of way the restriction happen and remove from the total number of way to arrange the n !


It is 2 ! for arrangement and 4 ! for the remaining 4 people

but what I don t understand is why it is time by 5 as the OA gives


I saw some other type like that

For instance digit 1,2,3,4,5 IF EACH DIGIT is used only once how many ways can each digit be arranged such 2 and 4 are not adjacent -
In this case the restriction is 2!x4! not multiplied by anything else
Can anyone explain me why


thanks for your Time


regards


Dear Moderator,
How can it be A , when no choices are given, hope you will look into this. Thank you.
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Re: 6 persons are going to theater and will sit next to each oth  [#permalink]

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New post 07 Sep 2018, 11:07
Maude wrote:
6 persons are going to theater and will sit next to each other in 6 adjacent seats. But Martia and Jan can not sit next to each other. In how many arrangement can this be done

I understood that the restriction must be deal first
by finding the number of way the restriction happen and remove from the total number of way to arrange the n !


It is 2 ! for arrangement and 4 ! for the remaining 4 people

but what I don t understand is why it is time by 5 as the OA gives


I saw some other type like that

For instance digit 1,2,3,4,5 IF EACH DIGIT is used only once how many ways can each digit be arranged such 2 and 4 are not adjacent -
In this case the restriction is 2!x4! not multiplied by anything else
Can anyone explain me why


thanks for your Time


regards


Total number of arranging 6 persons \(=6!\)
Total number of arranging 6 persons such that Matina and Jan are together \(=2*5!\)
the number of arranging 6 persons such that Matina and Jan are not together \(= 6!-2*5!\)
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Re: 6 persons are going to theater and will sit next to each oth   [#permalink] 07 Sep 2018, 11:07
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