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Mathematics 14 Online
OpenStudy (anonymous):

evaluate 15 C 4 and 15 p 4

OpenStudy (anonymous):

do u understand it

OpenStudy (anonymous):

No.

OpenStudy (anonymous):

What's C and p?

OpenStudy (anonymous):

Its part of the formula

OpenStudy (anonymous):

they are combinations and permuatations

OpenStudy (anonymous):

nCr = n! / (n-r)!r!

OpenStudy (anonymous):

nPr = n! / (n-r)!

OpenStudy (anonymous):

I see. Never seen it before!!

OpenStudy (anonymous):

there are buttons on calculators for it

OpenStudy (anonymous):

15C4 = 1365 15P4 = 32760

OpenStudy (anonymous):

conclusion , there are alot more arrangements if you consider order !

OpenStudy (anonymous):

so then u devide them

OpenStudy (anonymous):

its combinatorics , a bit of a painful topic, a bit tricky sometimes

OpenStudy (anonymous):

what u do with those 2 now?

OpenStudy (anonymous):

what do you mean "what do you do with them now" , you asked the question :|

OpenStudy (anonymous):

yeah but thats not the answer

OpenStudy (anonymous):

its only suppose to be one answer because ur evaluation

OpenStudy (anonymous):

:|

OpenStudy (anonymous):

so you meant 15C4 + 15P4 :|

OpenStudy (anonymous):

i said and

OpenStudy (anonymous):

yeh :|

OpenStudy (anonymous):

would it be 24

OpenStudy (anonymous):

you could possibly be referring to the multiplication of them , but thats really badly worded

OpenStudy (anonymous):

and is add " of" is mutiplication

OpenStudy (anonymous):

that was right can u help me with the one I just posted up sorry

OpenStudy (anonymous):

Its the same stuff just diff question

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