Question

This is all the question gave me 6 C 2 = _______ a0

Answer 1

15

Explanation:

Using the definition of n
`C_(r)`

n
`C_(r)` =
`(n!)/(r!(n-r)!)`

where n! = n(n - 1)(n - 2) × 3 × 2 × 1

Hence

6
`C_(2)`

=
`(6!)/(2!(4!))`

=
`(6(5)(4)(3)(2)(1))/(2(1)4(3)(2)(1))`

Cancel 4(3)(2)(1) on numerator/ denominator, leaving

=
`(6(5))/(2(1))` =
`(30)/(2)` = 15