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} $