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

$\\ \rm\longmapsto ^nC_r=\dfrac{n!}{r!(n-r)!}$

$\\ \rm\longmapsto ^{12}C_3$

$\\ \rm\longmapsto \dfrac{12!}{3!(12-3)!}$

$\\ \rm\longmapsto \dfrac{12!}{3!(9!)}$

$\\ \rm\longmapsto \dfrac{12(11)(10)(9!)}{3!(9!)}$

• cancel 9!

$\\ \rm\longmapsto \dfrac{12(11)(10)}{3!}$

$\\ \rm\longmapsto \dfrac{1320}{3(2)(1)}$

$\\ \rm\longmapsto \dfrac{1320}{6}$

$\\ \rm\longmapsto 220$