Eight friends Bertha, Beretta, Bessy, Buena, Greta, and Olga are sitting in a row for a group

Eight friends Bertha, Beretta, Bessy, Buena, Greta, and Olga are sitting in a row for a group
Photo. Whit is the probability that Bertha and olga are not sitting together?

i dont know

...

fucking combinatorics man
1-8C2/8! ????

Bertha could be in positions 2 through 7 and have Olga on either side. 12 possibilities.
Bertha could be in position 1 or 8 and have Olga next to her. 2 possibilities.
14 in all.
8 people can be ordered 8! ways. 40320.
Probability they're adjacent 14/40320
Probability they're not adjacent 1-14/40320

>8 friends
>lists 6 people

Still waiting for those last two, user.
Fucking math kid can't even count in N.

0%, because all of them are sitting together for the photo!

You only gave 6 names.
We have to assume there aren't 20 people in that row, 8 of whom are friends.

I find it hard to believe the chance of them sitting together is that low

This was confusing me soo much

75%

Haha my bad Beatrice and beonka

Correct

wrong

Care to explain?

Yes please explain

100% given that they are BFFs.

jeeze I thought people on this board were smart
it's like the other user said, take the position of one friend along the row. there are two cases: an end position (2), in which the other friend has one possible seat, and a middle position (6) where the second friend has two options. so the choices for adjacent seating are (2*6!+6*2*6!)/8!=1/4 and their complement is 3/4

there are 6! * 7 * 2 = 10080 different ways these 2 girls can sit next to each other which 25% of the total combination. so 1-.25=.75

you're counting positions multiple times

I see. They can be 1-2, 2-3, 3-4, etc.
But in each case the other six girls can be arranged in numerous ways.

nope, think carefully. I'm counting all appropriate positions once and only once by iterating the position of first one girl and then the other, and then accounting for the 6! possible positions of the other girls in the remaining seats. this is easier to think about than many combinatorics problems because order matters.

why would order matter?

an arrangement of 8 distinct objects in a row does not have any symmetries. if you take a photograph of 8 people in a line and rotate it 180 degrees or mirror it across a vertical centre line it will be a different photo.

1:3

3/4
Let's count the possibilities, where they are sitting together. The easiest way is that we count the 2 of them as 1, because they are always sitting together, so we only have 7 people now, and they can sit in 7! possible different ways. The 2 girls can switch places so it's actually 2*7!. Now if we divide this by all of the possibilities the 8 people can sit, we get 1/4 (2*7!/8!=2/8=1/4). So the possibility they are sitting together is 1/4 therefore the possibility of them not sitting together is 1-1/4=3/4. The amount of people who got this wrong shows, how many brainlets are larping here as "math experts".

Btw this is basic high school level combinatorics, that an average 16 years old should have no problem with

I wasn't taught this in high school because I was on the accelerated track and skipped out of "basic" math. It's much harder to solve this kind of problem if you don't have experience.

>being so full of shit