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A, B, C, D, E, and F (representing people) go to a movie and [#permalink]
03 Dec 2003, 14:20

A, B, C, D, E, and F (representing people) go to a movie and want to sit in a row of 6 chairs, but B and D refuse to sit next to each other. How many different arrangements can they sit?

I took 6! and then made a chart to figure out how many arrangments there are with B and D together. Then I figured out the permutaion for the remaining 4 seats ( 4!) and then mutilipled that by the number of seating options with B and D together and then subtracted that number from 720. I think I got the correct answer, but this seems very labor intensive. Is there any easier way to do it? Thanks!

there is a 1/3 chance (240/720) that B is on either end, and there is a 1/5 chance (given that B is on the end) of E sitting next to B. That's 48 arrangements where they would be together.

there is a 2/3 chance (480/720) that B is not on either end. If b is in an interior seat, there is a 2/5 chance that E will sit next to him. That's 192 arrangements where they would sit together.

240/720 arrangements put them together.
480/720 keep them apart.

Re: permutation problem [#permalink]
03 Dec 2003, 15:09

mbartow wrote:

A, B, C, D, E, and F (representing people) go to a movie and want to sit in a row of 6 chairs, but B and D refuse to sit next to each other. How many different arrangements can they sit?

1. total # of ways to seat 6 people in a row = 6! = 720
2. total # of ways in which B & D *can* sit together = 5! * 2! = 240
Consider B and D as one single unit , thats 5 people now...also, these
two can themselves be arranged in 2! ways ..hence 240 total ways
3. Required # = 720 - 240 = 480 ways

Please correct my assumptions if I am wrong. I'm trying to understand the concept behind this problem.

6! is straight forward for a total number of possibilities.
= 720

I understand the 5!; you're treating two people as one which in effect still fills 6 seats b/c the two jokers will be sitting next to each other using 5!
= 120

2! comes into play b/c order matters w/our two guys
a,b is not equal to b,a and thus we have to account for both arrangements

Do I have it straight?

Now, let's take this example a step farther.
If we have 7 people in a movie theater, and 3 people choose not to sit next to each other, how many arrangements do we have?

7! = total possibilities = 5040

To fill all 7 seats, we take 5!, b/c 1 unit is now 3
5! = 120

120 * 3! b/c there are 6 different possibilities for 3 people to be seated.
120 * 6 = 720 arrangements where these 3 jokers sit next to each other

I think you have figured it out. One way of thinking of these problems is to treat them as a series of "event possibilities". When you combine 2 people and treat them as one unit, you figure out one of the possibilities, as you did with 5! or 6!.

Once you have figured out one set of possibilities, figure out the next set of possibilities, which is 2! (arrangement of 2 people) or 3! (arrangement of 3 people).

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