Question #7c757
1 Answer
Feb 2, 2017
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#
