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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] New post 16 Jun 2010, 19:08
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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

[Reveal] Spoiler:
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
[Reveal] Spoiler: OA
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Re: Permutations with Restriction [#permalink] New post 16 Jun 2010, 19:45
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] New post 16 Jun 2010, 19:54
Expert's post
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: Permutations with Restriction   [#permalink] 16 Jun 2010, 19:54
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6 persons are going to theater and will sit next to each oth

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