JEE Main & Advanced Mathematics Permutations and Combinations Question Bank Mock Test - Permutations and Combinations

  • question_answer
    Total number of six-digit numbers that can be formed having the property that every succeeding digit is greater than the preceding digit is equal to

    A) \[^{9}{{C}_{3}}\]                 

    B) \[^{10}{{C}_{3}}\]

    C) \[^{9}{{P}_{3}}\]                  

    D) \[^{10}{{P}_{3}}\]

    Correct Answer: A

    Solution :

    [a]\[{{x}_{1}}<{{x}_{2}}<{{x}_{3}}<{{x}_{4}}<{{x}_{5}}<{{x}_{6}},\]when the number Is \[{{x}_{1}}{{x}_{2}}{{x}_{3}}{{x}_{4}}{{x}_{5}}{{x}_{6}}\]clearly no digit can be zero. Also, all the digits are distinct. So. Let us first select six digits from the list of digits 1, 2, 3, 4, 5, 6, 7, 8, 9, which can be done in \[^{9}{{C}_{6}}\] ways. After selecting these digits they can be put only in one order. Thus, total number of such numbers is \[^{9}{{C}_{6}}\times 1{{=}^{9}}{{C}_{3.}}\]

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