# [Probability] Seating arrangement & Selection

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:

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

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

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
-------------------------------------------------------------------------------------------------------
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:
7! - (6! x 2) = 3600
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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
7! - (5! x 2) = 4800
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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 = 1/2
Different colors: 10/16 x 6/15 + 6/16 x 10/15 = 1/2

LCKurtz
Homework Helper
Gold Member
Thanks LCKurtz, OK, here's what I got:

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
7! - (5! x 2) = 4800
Remember the total number of ways 7 people can be arranged around a table is 6!, not 7!
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 = 1/2
Different colors: 10/16 x 6/15 + 6/16 x 10/15 = 1/2
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.

Remember the total number of ways 7 people can be arranged around a table is 6!, not 7!

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

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

thanks

LCKurtz
Homework Helper
Gold Member
2 red marbles = 2 x 10 = 20
2 blue marbles = 2 x 6 = 12
No. Have you studied permutations and combinations? Binomial coefficients? Do these symbols mean anything to you?

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

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!

LCKurtz