contestada

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.

Respuesta :

[tex]\\ \rm\longmapsto ^nC_r=\dfrac{n!}{r!(n-r)!}[/tex]

[tex]\\ \rm\longmapsto ^{12}C_3[/tex]

[tex]\\ \rm\longmapsto \dfrac{12!}{3!(12-3)!}[/tex]

[tex]\\ \rm\longmapsto \dfrac{12!}{3!(9!)}[/tex]

[tex]\\ \rm\longmapsto \dfrac{12(11)(10)(9!)}{3!(9!)}[/tex]

  • cancel 9!

[tex]\\ \rm\longmapsto \dfrac{12(11)(10)}{3!}[/tex]

[tex]\\ \rm\longmapsto \dfrac{1320}{3(2)(1)}[/tex]

[tex]\\ \rm\longmapsto \dfrac{1320}{6}[/tex]

[tex]\\ \rm\longmapsto 220[/tex]

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