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Permutations and Combinations question

2002 · Shift 0 · Q107
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Permutations and Combinations question

2002 · Shift 0 · Q107

JEE MainMathematicsPermutations and CombinationsMCQ+4 / −1
Number greater than 1000 but less than 4000 is formed using the digits 0, 1, 2, 3, 4 (repetition allowed). Their number is :
  1. A
    125
  2. B
    105
  3. C
    374
  4. D
    625
View written solutionFree

Correct answer: C

  1. We need to form a number:

    • greater than 100010001000
    • less than 400040004000
    • using digits 0,1,2,3,40,1,2,3,40,1,2,3,4
    • repetition allowed.
  2. Since the number is between 100010001000 and 400040004000, it must be a 4-digit number.

  3. The thousands digit determines the range:

    • It cannot be 000 (otherwise not a 4-digit number).
    • It cannot be 444, because then the number would be at least 400040004000, which is not allowed.
    • So the thousands digit can only be 1,2,31,2,31,2,3.

    Hence, number of choices for the first digit = 333.

  4. For each of the remaining three places (hundreds, tens, units):

    • any of the digits 0,1,2,3,40,1,2,3,40,1,2,3,4 can be used,
    • repetition is allowed.

    So each of these places has 555 choices.

  5. Total number of such numbers: 3×5×5×5=3×125=375.3 \times 5 \times 5 \times 5 = 3 \times 125 = 375.3×5×5×5=3×125=375.

  6. Check the boundary condition:

    • 100010001000 itself is included in this count, but the question says greater than 100010001000, so exclude 100010001000.
    • 400040004000 is not included anyway, since leading digit 444 was not allowed.

    Therefore required count is: 375−1=374.375 - 1 = 374.375−1=374.

  7. Hence the correct option is: 374\boxed{374}374​ which is Option C.

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