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Mathematics 9 Online
OpenStudy (anonymous):

An administrative assistant has 4 blue file folders, 3 red folders, and 3 yellow folders on her desk. Each folder contains different information, so two folders of the same color should be viewed as being different. She puts the file folders randomly in a box to be taken to a meeting. Find each probability. 1. P(4 blue, 3 red, 3 yellow, in that order) 2. P(first 2 blue, last 2 blue)

OpenStudy (zarkon):

where are you stuck?

OpenStudy (anonymous):

all

OpenStudy (zarkon):

do you know how many ways you can arrange the 10 folders?

OpenStudy (anonymous):

1) 4!3!3! ?

OpenStudy (zarkon):

that is actually the numerator to the first problem :)

OpenStudy (anonymous):

1) 4!3!3! M'I corect ?

OpenStudy (zarkon):

what is the denominator?

OpenStudy (anonymous):

10

OpenStudy (zarkon):

10!

OpenStudy (zarkon):

\[\frac{4!3!3!}{10!}\]

OpenStudy (anonymous):

for frist one?

OpenStudy (zarkon):

yes

OpenStudy (anonymous):

how second?

OpenStudy (zarkon):

same experiment so 10! is still the denominator

OpenStudy (zarkon):

you have 4 choices for the first blue,3 for the second 2 for the second to last spot and 1 for the very last spot and 6! for the middle so 4!6! \[\frac{4!6!}{10!}\]

OpenStudy (anonymous):

great ,Thank you

OpenStudy (zarkon):

np

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