How do you evaluate #""^12C_6#?
1 Answer
Jan 27, 2017
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