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JEE Main & Advanced Mathematics Permutations and Combinations Question Bank Multinomial Theorem, Number of Divisors, Miscellaneous problems

  • question_answer
    If \[P(n,r)=1680\] and \[C(n,r)=70\], then \[69n+r!=\] [Kerala (Engg.)2005]

    A) 128

    B) 576

    C) 256

    D) 625

    E) 1152

    Correct Answer: B

    Solution :

    \[P(n,r)=1680\]\[\frac{n!}{(n-r)!}=1680\] ?..(i) \[C(n,r)=70\]Þ\[\frac{n!}{r!\,(\,n-r)!}=70\] ?..(ii) \[\frac{1680}{r!}=70\], [From (i) and (ii)] \[r!=\frac{1680}{70}=24\]Þ \[r=4\] \[\because P(n,\,4)=1680\] \[\because \]\[n(n-1)(n-2)(n-3)=1680\] Þ \[n=8\] \[\because \] \[8\times 7\times 6\times 5=1680\] Now \[69n+r\,!=69\times 8+4!\]\[=552+24\]= 576.


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