Question

$\frac{{}^8{C}_0}{6}{-}^8{C}_1{+}^8{C}_2\cdot 6{-}^8{C}_3{.6}^2+....{+}^8{C}_8{\cdot 6}^7$ is equal to

• A. $0$
• B. $6^7$
• C. $6^8$
• D. $\frac{5^8}{6}$

Answer 1

Answer: (D)

$\frac{{}^8{C}_0}{6}{-}^8{C}_1{+}^8{C}_2.6{-}^8{C}_3{6}^2+....{+}^8{C}_8{6}^7$
$=\frac{1}{6}{[}^3{C}_0-6^8{C}_1+6^2{}^8{C}_2-6^3{}^8{C}_3+....+6^8{}^6{C}_8]$
$=\frac{1}{6}[{(1-6)}^8]=\frac{5^8}{6}$