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OpenStudy (anonymous):
OpenStudy (anonymous):
@iGreen
OpenStudy (anonymous):
Could you help me maybe? :)
OpenStudy (anonymous):
@confluxepic Do you think you could help?
OpenStudy (igreen):
Hmm..I think it's going to be \(35 \times 34 \times 33 \times 32\)..
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OpenStudy (igreen):
For the first question ^^
OpenStudy (igreen):
@hartnn
OpenStudy (anonymous):
So for the first one it would be c?
OpenStudy (igreen):
I think..I'm not sure.
hartnn (hartnn):
35 C 4
or
35*34*33*32
is correct :)
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OpenStudy (igreen):
Oh, okay. Thanks @hartnn
So yes, C is correct.. @Pamela16
OpenStudy (anonymous):
Thank you! do you think you could help with the last one
OpenStudy (igreen):
I can try.
OpenStudy (anonymous):
Take your time :)
hartnn (hartnn):
i'll let iGreen try :)
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OpenStudy (anonymous):
Okay
OpenStudy (igreen):
Well, if all the letters were different we could just go \(5 \times 4 \times 3 \times 2 \times 1\)..but there are two E's here so if we did \(5 \times 4 \times 3 \times 2 \times 1\) we will have combinations that are the same.
hartnn (hartnn):
good start :)
OpenStudy (anonymous):
Okay so that would = 120
OpenStudy (igreen):
No, I just said we can't do that.. out of those 120 combinations there will be some that are the same because we have two of the same letter..E.
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OpenStudy (anonymous):
Oh okay I apoligize
OpenStudy (igreen):
Okay, I'll just leave this to @hartnn to solve..lol. I have no idea where to go from there.
hartnn (hartnn):
and then how would you eliminate duplicates ?
OpenStudy (igreen):
I don't know. ^^
OpenStudy (igreen):
I never learned how to do this stuff..lol.
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OpenStudy (anonymous):
Oh okay :) lol
hartnn (hartnn):
let me give you examples, so that you get it easily :)
abcd ---> 4!
aabc ----> 4!/2!
abcde --->5!
aabcd --->5!/2!
aaabc---->5!/3!
aaabb ---->5!/(3!2!)
see the pattern ? :)
OpenStudy (igreen):
Kind of..
OpenStudy (anonymous):
Im honestly thinking its a or c im maybe wrong?
hartnn (hartnn):
cheer is like
aabcd
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