A, B, C, D, E, and F (representing people) go to a movie and

praet,

could you show me how to do step 2 using Combinations, i can't see it...

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

... so ...

5040 - 720 = 4320 total possibilities

Is this correct?

Thanks,
CJ

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).

Lastly, multiply the 2 : 5! * 2!....