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)

$\frac{{{5}^{8}}}{6}$

Answer 2

$\frac{^{8}{{C}_{0}}}{6}{{-}^{8}}{{C}_{1}}{{+}^{8}}{{C}_{2}}.6{{-}^{8}}{{C}_{3}}{{6}^{2}}+....{{+}^{8}}{{C}_{8}}{{6}^{7}}$
$=\frac{1}{6}{{[}^{3}}{{C}_{0}}-{{6}^{8}}{{C}_{1}}+{{6}^{2}}{{\,}^{8}}{{C}_{2}}-{{6}^{3}}{{\,}^{8}}{{C}_{3}}+....+{{6}^{8}}{{\,}^{6}}{{C}_{8}}]$
$=\frac{1}{6}[{{(1-6)}^{8}}]=\frac{{{5}^{8}}}{6}$