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Mathematics 24 Online
OpenStudy (anonymous):

Picture in attachment please check it out!

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 \(35 \times 34 \times 33 \times 32\)..

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 :)

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 :)

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.

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.

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

OpenStudy (igreen):

Ohhh!

OpenStudy (igreen):

And for aabcd we use 5!/2!..

OpenStudy (igreen):

So that's \(\dfrac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1}\)?

OpenStudy (igreen):

@hartnn

OpenStudy (anonymous):

My session is about to run out could someone tell me the answer?

OpenStudy (igreen):

I believe that's it.. just multiply 5 x 4 x 3 x 2 x 1, and then divide it by 2 x 1.

OpenStudy (igreen):

You already found out that 5 x 4 x 3 x 2 x 1 is 120..so divide that by 2..what do you get? @Pamela16

OpenStudy (anonymous):

60

OpenStudy (igreen):

Yep!

OpenStudy (igreen):

I think that's your answer.

hartnn (hartnn):

60 it is :)

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