Question

$\left({}^8{C}_1{-}^8{C}_2{+}^8{C}_3{-}^8{C}_4{+}^8{C}_5{-}^8{C}_6{+}^8{C}_7{-}^8{C}_8\right)$ equals:

• A. 0
• B. 1
• C. 70
• D. 256

Answer 1

Answer: (B)

Let $A=\left({}^8{C}_1{-}^8{C}_2{+}^8{C}_3{-}^8{C}_4{+}^8{C}_5{-}^8{C}_6{+}^8{C}_7{-}^8{C}_8\right)$
$=\frac{8!}{1!7!}-\frac{8!}{2!6!}+\frac{8!}{3!5!}-\frac{8!}{4!4!}+\frac{8!}{5!3!}-\frac{8!}{6!2!}+\frac{8!}{7!1!}-\frac{8!}{0!8!}$
Note : ${}^n{C}_r=\frac{n!}{r!\left(n-r\right)!}$
Thus, $A=8-\frac{8\times 7}{2}+\frac{8\times 7\times 6}{3\times 2}-\frac{8\times 7\times 6\times 5}{4\times 3\times 2\times 1}+\frac{8\times 7\times 6}{3\times 2}-\frac{8\times 7}{2}+8-1$
And A = 8 - 28 + 56 - 70 + 56 - 28 + 8 - 1 = 1