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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2002

  • question_answer
    How many nine digit numbers can be formed using the digits 2, 2, 3, 3, 5,5, 8, 8, 8 so that the odd digits occupy even positions:

    A)  180                                       

    B)  7560

    C)  60                                         

    D)  16

    Correct Answer: C

    Solution :

    4 odd digits in 4 even places can be arranged in 4! Ways. But 2 alike \[(3,3),2\]alike \[(5,5)\]are there \[\therefore \]Number of ways \[=\frac{4!}{2!2!}=\frac{24}{2}=6\] ?(i) Remaining 5 even digits (2 alike, 3 alike) in 5 places can be arranged as above in \[\frac{5!}{2!3!}=\frac{120}{2\times 6}=10\,\text{ways}\]                             ?(2) By fundamental theorem real number \[=6\times 10=60\]


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