Question

Given 12 consecutive integers, how many ways can three of these integers be selected to give a sum which divides by 4.

Disclaimer: A lot of points to be given, Full explanation required. Not only answer.

Answer 1

`\\ \rm\longmapsto ^nC_r=(n!)/(r!(n-r)!)`

`\\ \rm\longmapsto ^(12)C_3`

`\\ \rm\longmapsto (12!)/(3!(12-3)!)`

`\\ \rm\longmapsto (12!)/(3!(9!))`

`\\ \rm\longmapsto (12(11)(10)(9!))/(3!(9!))`

• cancel 9!

`\\ \rm\longmapsto (12(11)(10))/(3!)`

`\\ \rm\longmapsto (1320)/(3(2)(1))`

`\\ \rm\longmapsto (1320)/(6)`

`\\ \rm\longmapsto 220`