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Mathematics 7 Online
OpenStudy (anonymous):

Asa, Bill, Cathy, Dino, and Eve are randomly seated in a row of 5 seats. Find the probability that Asa will sit next to Bill.

OpenStudy (anonymous):

View asa and bill as one person. Then, find the number of ways the four "people" can fill the seats. Then view bill and asa as one person, and do it again. Sum the two. Thats the total number of ways bill and asa can sit next to each other. Now, find the total number of ways the five students can sit in the row. The probability they sit next to each other is the the number of ways they can sit next to each other divided by the number of ways they can sit overall.

OpenStudy (anonymous):

Note that order is important!

OpenStudy (anonymous):

can u show the work?? lol

OpenStudy (anonymous):

Can you?

OpenStudy (anonymous):

9 / 5!

OpenStudy (anonymous):

8 / 5!

OpenStudy (anonymous):

what? It's supposed to be a fraction

OpenStudy (anonymous):

you have ABCDE , there are 5! ways of sitting them. that will be your denominator

OpenStudy (anonymous):

the favorable is AB sitting together, you can have oh...

OpenStudy (anonymous):

you can have AB, BA , and there are 4 places , so 4* 2

OpenStudy (anonymous):

thats your numerator

OpenStudy (anonymous):

what's the numerator?

OpenStudy (anonymous):

8

OpenStudy (anonymous):

so you have 8/120

OpenStudy (anonymous):

that's not a choice though (1) 1/60 (2) 2/5 (3) 1/10 (4) 4/15

OpenStudy (anonymous):

oh

OpenStudy (anonymous):

ok there are 2x4x3! in the numerator

OpenStudy (anonymous):

so 2/5

OpenStudy (anonymous):

wait 24/120 is 1/5 not 2/5

OpenStudy (anonymous):

no thats 48/120

OpenStudy (anonymous):

oh, I got u

OpenStudy (anonymous):

2x4x3! = 2x4x(3x2x1)

OpenStudy (anonymous):

give me medal NOW

OpenStudy (anonymous):

I got 2/5.

OpenStudy (anonymous):

me too

OpenStudy (anonymous):

there are 2 ways for AB to sit next to each other, they can move around in 4 different places, then there are 3! for the rest of the other people to sit

OpenStudy (anonymous):

wait so basically it's 5!/ 2!4! ???

OpenStudy (anonymous):

shut up omg. Look at the method i proposed. 2(4!)/(5!)

OpenStudy (anonymous):

well you treat AB as K, so you have K C D E , and how many ways can you reararnge that

OpenStudy (anonymous):

then K has 2! ways

OpenStudy (anonymous):

shut up, omg, GROW UP

OpenStudy (anonymous):

u 2 r arguing?

OpenStudy (anonymous):

so you have K C D E , where K is AB, and you have 4! ways of rearranging that, but K is actually AB or BA, so 4! * 2!

OpenStudy (anonymous):

azn, can you give me medal, i got answer first

OpenStudy (anonymous):

can u guys explain what u did?

OpenStudy (anonymous):

i did

OpenStudy (anonymous):

explain then right now

OpenStudy (anonymous):

treat AB as K, since they always sit next to each other, so you have K C D E , correct ?

OpenStudy (anonymous):

yea

OpenStudy (anonymous):

so how many ways to rearrange K C D E ?

OpenStudy (anonymous):

24

OpenStudy (anonymous):

right, but for each one of those arrangements, you have AB or BA

OpenStudy (anonymous):

since they can swap

OpenStudy (anonymous):

so you need 24 x 2

OpenStudy (anonymous):

in other words, K can be AB or K can be BA

OpenStudy (anonymous):

so you have 24 with AB , and 24 with BA, so 24 + 24

OpenStudy (anonymous):

ok and how did u get the 5! part

OpenStudy (anonymous):

thats the number of ways to rearrange A B C D E

OpenStudy (anonymous):

5 !

OpenStudy (anonymous):

wheres my medal

OpenStudy (anonymous):

PLEASSSSSSSSEEEEEEEEEEEEEEEEEEEEEEEEE

OpenStudy (anonymous):

oh I got it

OpenStudy (anonymous):

how do I give u a medal? It's my first time here, how?

OpenStudy (anonymous):

next to my name

OpenStudy (anonymous):

it should say give medal

OpenStudy (anonymous):

it doesn't

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