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Manipal Engineering Manipal Engineering Solved Paper-2013

  • question_answer
    A person is permitted to select at least one and at most \[n\] coins from a collection\[(2n+1)\]distinct coins. If the total number of ways in which he can select coins is 255, then \[n\] equals

    A)  4                                            

    B)  8

    C)  16                         

    D)         32

    Correct Answer: A

    Solution :

    We have \[^{2n+1}{{C}_{1}}{{+}^{2n+1}}{{C}_{2}}+...{{+}^{2n+1}}{{C}_{n}}=255\] ? (i) Also the sum of binomial coefficients \[^{2n+1}{{C}_{0}}{{+}^{2n+1}}{{C}_{1}}{{+}^{2n+1}}{{C}_{2}}+...{{+}^{2n+1}}{{C}_{n}}\] \[{{+}^{2n+1}}{{C}_{n+1}}+{{+}^{2n+1}}{{C}_{n+2}}+...{{+}^{2n+1}}{{C}_{2n+1}}\] \[={{(1+1)}^{2n+1}}={{2}^{2n+1}}\] \[\Rightarrow \]\[^{2n+1}{{C}_{0}}+2{{\{}^{2n+1}}{{C}_{1}}{{+}^{2n+1}}{{C}_{2}}+...{{+}^{2n+1}}{{C}_{n}}\}\]                                                 \[{{+}^{2n+1}}{{C}_{2n+1}}={{2}^{2n+1}}\] \[\Rightarrow \]\[1+2(255)+1={{2}^{2n+1}}\] \[\Rightarrow \]               \[1+255={{2}^{2n}}\] \[\Rightarrow \]                     \[{{2}^{2n}}={{2}^{8}}\] \[\Rightarrow \]                         \[n=4\]


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