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Mathematics 21 Online
OpenStudy (anonymous):

Find the binomail coefficient: (2012) (2011)

OpenStudy (anonymous):

@FibonacciChick666

OpenStudy (anonymous):

@Nnesha

OpenStudy (anonymous):

@AnswerMyQuestions

OpenStudy (anonymous):

HELP PLZZZZZZZZZ

OpenStudy (fibonaccichick666):

do you mean convert those numbers to binomial form?

OpenStudy (anonymous):

yes I think its one of these 2011, 0, 2012, or 1

OpenStudy (anonymous):

im not sure though.. but do u know how to do this

OpenStudy (anonymous):

@DuckDynastyfan923

OpenStudy (anonymous):

the answers are A. 2011

OpenStudy (anonymous):

B. 0

OpenStudy (anonymous):

C. 2012

OpenStudy (anonymous):

D. 1

OpenStudy (anonymous):

Also no its just says "find the binomail coefficient: (2012) (2011)

OpenStudy (anonymous):

@Abhisar

OpenStudy (anonymous):

how r u working it

OpenStudy (anonymous):

@Legends

OpenStudy (anonymous):

so?

OpenStudy (anonymous):

should i get us help

OpenStudy (anonymous):

HELLPPPPp

OpenStudy (anonymous):

HELPPPPPPPPPP USSSSSS

OpenStudy (mathmate):

Do you mean: \[\left(\begin{matrix}2012 \\ 2011\end{matrix}\right)\] ?

OpenStudy (anonymous):

yes

OpenStudy (mathmate):

A binomial coefficient \(\left(\begin{matrix}n \\ r\end{matrix}\right)\) is defined as \(\dfrac{n!}{(n-r)!r!}\) So substitute the numbers and find your answer.

OpenStudy (anonymous):

yes but after that wat

OpenStudy (anonymous):

i can do that its just i dont know what to do after

OpenStudy (mathmate):

You will need to substitute n=2012, r=2011, and calculate your answer, choose from the choices.

OpenStudy (anonymous):

ok so D then considering they be nothing i else i tihnk

OpenStudy (anonymous):

im thinking d

OpenStudy (anonymous):

cant be 2011 or 2012

OpenStudy (mathmate):

It's ok with me if you decide to guess and not calculate!

OpenStudy (anonymous):

give me one sec

OpenStudy (anonymous):

@mathmate i did it but then wat

OpenStudy (anonymous):

what do you do with the extra 2011

OpenStudy (anonymous):

HELP!!!

OpenStudy (amoodarya):

\[\left(\begin{matrix}n \\ k\end{matrix}\right)=\left(\begin{matrix}n \\n- k\end{matrix}\right)\\so\\\left(\begin{matrix}2012 \\2011\end{matrix}\right)=\left(\begin{matrix}2012 \\2012-2011\end{matrix}\right)=\\ \left(\begin{matrix}2012 \\ 1\end{matrix}\right)=2012\]

OpenStudy (anonymous):

really 20212 i thought 1

OpenStudy (amoodarya):

\[\left(\begin{matrix}2012 \\ 1\end{matrix}\right)=\frac{2012!}{1! *2011!}=\\\frac{1*2*3...2011*2012}{1! (1*2*3...2011)=2012}\]

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