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