# [Probability] Seating arrangement & Selection

1. Nov 20, 2009

### CountNumberla

PROBLEM 1
7 people sit in a row, and 2 of them want to sit together, in how many ways can they be arranged?

ATTEMPTED SOLUTION:
6! x 2 = 1440
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PROBLEM 2
7 people sit in a row, and 2 of them do not want to sit together, in how many ways can they be arranged?

ATTEMPTED SOLUTION:
No ideas
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PROBLEM 3
7 people sit in a circular formation, but 2 of them do NOT want to sit together, in how many ways can they be arranged?

ATTEMPTED SOLUTION
No ideas
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PROBLEM 4
There are 10 red & 6 blue marbles in a bag. You pick 2 of them at the same time.
In how many ways can you pick 2 marbles of the same color? Different colors?

ATTEMPTED SOLUTION:
Same colors: 10/16 x 9/15 + 6/16 x 5/15
Different colors: 10/16 x 6/16 + 9/15 x 5/15

Last edited: Nov 20, 2009
2. Nov 20, 2009

### LCKurtz

Looks good.
How about subtracting out the ones where they sit together from the total?
Circular permutations, same idea as 1.
Think about combinations. How many ways can you choose 2 reds? 2 blues? So...

3. Nov 21, 2009

### CountNumberla

Thanks LCKurtz, OK, here's what I got:

4. Nov 21, 2009

### LCKurtz

Remember the total number of ways 7 people can be arranged around a table is 6!, not 7!
Hopefully you know that a fraction can't be the answer to how many ways to choose. I will ask you again: In how many ways can you pick 2 red marbles? 2 blue marbles? Think about combination formulas.

5. Nov 22, 2009

### CountNumberla

So it's safe to say, 6! - (5! x 2) = 480

Hmmmm
2 red marbles = 2 x 10 = 20
2 blue marbles = 2 x 6 = 12

thanks

6. Nov 22, 2009

### LCKurtz

No. Have you studied permutations and combinations? Binomial coefficients? Do these symbols mean anything to you?

$$C(n,r) =\binom{n}{r} = \frac{n!}{r!(n-r)!}$$

7. Nov 22, 2009

### CountNumberla

some permutations yes, but not that formula. However, I know a similar one.

so back to the original problem, 10 red and 6 blue, in how many ways can you pick 2 of the same color?

is it 16! / 10! 6! = 56

I'm not sure how to compute picking different colors though....

And are the others correct? Could you please give me the answer so I can fully grasp the concept?! This is driving me crazy and I don't want to fall behind with other stuff. I learn better by analyzing a correct problem, if that makes sense.

Thanks!

8. Nov 22, 2009

### LCKurtz

Well, no, I won't just give you the answer. Surely the text you are using must have some discussion of how many ways you can choose r things from n things and what it has to do with the combination formulas.

9. Nov 23, 2009

### CountNumberla

The book we use is VERY vague, with only a few examples, and we've only gone as far as the equation I've shown you. I don't understand why you cant help me with ONE answer, it's not like I haven't stressed myself trying to find the solution!! I mean you rather I never figure it out?

10. Nov 23, 2009