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JEE Main & Advanced Sample Paper JEE Main Sample Paper-37

  • question_answer
    Number of permutations\[1,\,\,2,\,\,3,\,\,4,\,\,5,\,\,6,\,\,7,\,\,8\]and\[9\] taken all at a time are such that the digit. \[1\] appearing somewhere to the left of \[2,\,\,\,3\] appearing to the left of \[4\] and \[5\] somewhere to the left of \[6\], is \[(e.g.,\] \[815723946\] would be one such permutation)

    A) \[9.7!\]                         

    B) \[8!\]

    C) \[5!.4!\]                        

    D) \[8!.4!\]

    Correct Answer: A

    Solution :

     Number of digits are\[9\] Select \[2\] places for the digit \[1\] and \[2\] in \[^{9}{{C}_{2}}\] ways from the remaining \[7\] places select any two places for \[3\] and \[4\] in \[^{7}{{C}_{2}}\] ways and from the remaining \[5\] places select any two for \[5\] and \[6\] in \[^{5}{{C}_{2}}\] ways Now, the remaining 3 digits can be filled in \[3!\] ways \[\therefore \]Total ways\[{{=}^{9}}{{C}_{2}}{{\cdot }^{7}}{{C}_{2}}{{\cdot }^{5}}{{C}_{2}}\cdot 3!\]   \[=\frac{9!}{2!.7!}\cdot \frac{7!}{2!.5!}\cdot \frac{5!}{2!3!}=9.7!\]


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