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.
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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!
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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
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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
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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
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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
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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
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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?
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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 ?
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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
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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
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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
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