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There are 6 friends-- A,B,C,D,E,F. In how many ways can they [#permalink]
31 Jan 2012, 16:37

There are 6 friends-- A,B,C,D,E,F. In how many ways can they sit in a row if

1) A is just before B.

2) A is before B.

My understanding is that for 1) it is 5 factorial (club AB as one unit ) and for 2) it is (6 factorial / 2) as out of total chances half the chances A might be before B and half the chances A might not. As there is slight difference in terms of wording 1and 2 I want to know whether my understanding is correct. Could someone please clarify ?

Thanks in advance
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Re: There are 6 friends-- A,B,C,D,E,F. In how many ways can they [#permalink]
31 Jan 2012, 18:54

Yes, your understanding is correct. Clubbing A and B together, A will be just before B in (6-1)! or 5! arrangements.

However, to calculate the number of arrangements where A will be before B, we divide 6! by 2 as in exactly half the arrangements, A will be before B, and in the other half B will be before A.

Note also that the number of arrangements where A will be before B has to be greater than the number of arrangements where A is just before B. This is true (120 v/s 360).
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Re: There are 6 friends-- A,B,C,D,E,F. In how many ways can they [#permalink]
01 Feb 2012, 22:30

Expert's post

gpkk wrote:

There are 6 friends-- A,B,C,D,E,F. In how many ways can they sit in a row if

1) A is just before B.

2) A is before B.

My understanding is that for 1) it is 5 factorial (club AB as one unit ) and for 2) it is (6 factorial / 2) as out of total chances half the chances A might be before B and half the chances A might not. As there is slight difference in terms of wording 1and 2 I want to know whether my understanding is correct. Could someone please clarify ?

Thanks in advance

Yes, I have discussed this distinction in the following two posts:

The first post discusses questions similar to question 1 and the second one discusses questions similar to question 2. You might want to go through them to be clear about the concepts involved.
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