Question

Question #7c757

Answer 1

`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`