Question

How do you evaluate 9C3?

Answer 1

`84`

Explanation:

in general

`color(white)(x)^nC_r=(n!) /(r!(n-r)!`

`color(white)(x)^9C_3=(9!) /(3!(9-3)!`

`=(9!)/(3!6!)`

#=(9xx8xx7xxcancel(6!))/ (3!cancel(6!)#

`=(cancel(9)^3xxcancel(8)^4xx7)/(cancel(3)xxcancel(2)xx1)`

`=3xx4xx7=84`

Answer 2

`84`

See steps below;

Explanation:

`"Formular" rArr ^nC_r = (n!)/((n - r)! r!)`

Where.... `c = 9, r = 3`

`rArr ^9C_3 = (9!)/((9 - 3)!3!)`

`color(white)(xxxx) rArr (9!)/((9 - 3)!3!)`

`color(white)(xxxx) rArr (9!)/((6)!3!)`

`color(white)(xxxx) rArr (9!)/(6!3!)`

`color(white)(xxxx) rArr (9 xx 8 xx 7 xx 6 xx 5 xx 4 xx 3 xx 2 xx 1)/((6 xx 5 xx 4 xx 3 xx 2 xx 1) (3 xx 2 xx 1))`

`color(white)(xxxx) rArr (9 xx 8 xx 7 xx cancel6 xx cancel5 xx cancel4 xx cancel3 xx cancel2 xx cancel1)/((cancel6 xx cancel5 xx cancel4 xx cancel3 xx cancel2 xx cancel1) (3 xx 2 xx 1))`

`color(white)(xxxx) rArr (9 xx 8 xx 7)/(3 xx 2 xx 1)`

`color(white)(xxxx) rArr 504/6`

`color(white)(xxxx) rArr 84`