How do you evaluate #""^12C_6#?

1 Answer
Jan 27, 2017

Answer:

#""^12C_6 = 924#

Explanation:

The general formula for combinations is:

#""^nC_r = (n!)/(r!(n-r)!)#

So in our example:

#""^12C_6 = (12!)/(6!6!)#

#color(white)(""^12C_6) = (12xx11xx10xx9xx8xx7)/(6xx5xx4xx3xx2xx1)#

#color(white)(""^12C_6) = (2^6xx3^3xx5xx7xx11)/(2^4xx3^2xx5)#

#color(white)(""^12C_6) = 2^2xx3xx7xx11#

#color(white)(""^12C_6) = 924#

Or you can write out Pascal's triangle to the #13#th row and pick the middle term in that row...

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