J & K CET Engineering J and K - CET Engineering Solved Paper-2009

  • question_answer
    \[\frac{^{8}{{C}_{0}}}{6}{{-}^{8}}{{C}_{1}}{{+}^{8}}{{C}_{2}}.6{{-}^{8}}{{C}_{3}}{{.6}^{2}}+....{{+}^{8}}{{C}_{8}}{{.6}^{7}}\] is equal to

    A)  \[0\]

    B)  \[{{6}^{7}}\]

    C)  \[{{6}^{8}}\]

    D)  \[\frac{{{5}^{8}}}{6}\]

    Correct Answer: D

    Solution :

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


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