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Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and

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Re: Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and [#permalink]

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New post 02 Feb 2018, 11:06
GMAT100 wrote:
Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and sit next to each other in 6 adjacent seats in the front row of the theater. If Marcia and Jan will not sit next to each other, in how many different arrangements


We can use the following formula:

Total number of ways to arrange the 6 people = (number of arrangements whenMarcia sits next to Jan) + (number of arrangements when Marica does not sit next to Jan)

Let’s determine the number of arrangements in which Marcia sits next to Jan.

We can denote Greg, Marcia, Peter, Jan, Bobby and Cindy as G,M, P, J, B, and C, respectively.

If Marcia and Jan must sit together, we can consider them as one person [MJ]. For example, one seating arrangement could be [MJ][G][P][B][C]. Thus, the number of ways to arrange five people in a row is 5! = 120.

However, we must also account for the ways we can arrange Marcia and Jan, that is, either [MJ] or [JM]. Thus, there are 2! = 2 ways to arrange Marcia and Jan.

Therefore, the total number of seating arrangements is 120 x 2 = 240 if Marcia and Jan DO sit next to each other.

Since there are 6 people being arranged, the total number of possible arrangements is 6! = 720.

Thus, the number of arrangements when Marcia does NOT sit next to Jan is 720 - 240 = 480.
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Re: Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and   [#permalink] 02 Feb 2018, 11:06
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Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and

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