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  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2005

  • question_answer
    If\[P(n,r)=1680\]and\[C(n,r)=70,\]then \[69n+r!\]is equal to:

    A)  128       

    B)                         576

    C)  256                       

    D)         625

    E)  1152

    Correct Answer: B

    Solution :

    \[\because \] \[^{n}{{p}_{r}}=1680\] \[\Rightarrow \]               \[\frac{n!}{(n-r)!}=1680\]                                                 ...(i) and        \[^{n}{{C}_{r}}=70\] \[\Rightarrow \]               \[\frac{n!}{(n-r)!r!}=70\]                                              ?.(ii) From Eqs. (i) and (ii), we get                 \[r!=\frac{1680}{70}=24=4!\Rightarrow r=4\] On putting value of r in Eq. (i), we get                 \[\frac{n!}{(n-4)!}=1680\] \[\Rightarrow \]               \[n(n-1)(n-2)(n-3)=1680\] \[\Rightarrow \]               \[n(n-1)(n-2)(n-3)=8(8-1)\]                                                                 \[(8-2)(8-3)\] \[\Rightarrow \]               \[n=8\] \[\therefore \]  \[69n+r!=69\times 8+24\]                                                 \[=552+24+576\]


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