Question #7c757

1 Answer
Feb 2, 2017

Answer:

#165#

Explanation:

Using the definition for #""^nC_r#

#color(red)(bar(ul(|color(white)(2/2)color(black)(""^nC_r=(n!)/(r!(n-r)!))color(white)(2/2)|)))#

#"where "n! =n(n-1)(n-2)(n-3)xx.....xx1#

#rArr""^(11)C_3=(11!)/(3!(11-3)!)=(11!)/(3!8!)#

#=(11xx10xx9)/(3xx2xx1)#

Note that the remaining product of factors on the numerator#color(blue)" cancel"# with the 8! on the denominator.

#rArr""^(11)C_3=990/6=165#