JEE Main & Advanced Mathematics Permutations and Combinations Question Bank Critical Thinking Questions

  • question_answer A five digit number divisible by 3 has to formed using the numerals 0, 1, 2, 3, 4 and 5 without repetition. The total  number of ways in which this can be done is [IIT 1989; AIEEE 2002]

    A) 216

    B) 240

    C) 600

    D) 3125

    Correct Answer: A

    Solution :

    We know that a five digit number is divisible by 3, if and only if sum of its digits (= 15) is divisible by 3, therefore we should not use 0 or 3 while forming the five digit numbers. Now, (i) In case we do not use 0 the five digit number can be formed (from the digit 1, 2, 3, 4, 5) in \[^{5}{{P}_{5}}\]ways. (ii) In case we do not use 3, the five digit number can be formed (from the digit 0, 1, 2, 4, 5) in \[\ 6\ !\ -5\ !\ \times 2=480\] ways. \[\therefore \] The total number of such 5 digit number \[={{\,}^{5}}{{P}_{5}}+{{(}^{5}}{{P}_{5}}{{-}^{4}}{{P}_{4}})=120+96=216\].


You need to login to perform this action.
You will be redirected in 3 sec spinner