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Permutations and Combinations question

2023 · 25 Jan · Shift 2 · Q34
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  5. /2023 · 25 Jan · Shift 2 · Q34

Permutations and Combinations question

2023 · 25 Jan · Shift 2 · Q34

JEE MainMathematicsPermutations and CombinationsMCQ+4 / −1
The number of numbers, strictly between 5000 and 10000 can be formed using the digits 1, 3, 5, 7, 9 without repetition, is :
  1. A
    120
  2. B
    6
  3. C
    72
  4. D
    12
View written solutionFree

Correct answer: C

  1. We need numbers strictly between 500050005000 and 100001000010000 formed from the digits 1,3,5,7,91,3,5,7,91,3,5,7,9 without repetition.

  2. Any number between 500050005000 and 100001000010000 must be a 4-digit number.

  3. For a 4-digit number to be greater than 500050005000, its thousand's digit must be at least 555. From the given digits 1,3,5,7,91,3,5,7,91,3,5,7,9, the possible thousand's digits are: 5,7,95,7,95,7,9 So, number of choices for the first digit =3=3=3.

  4. After choosing the first digit, we must fill the remaining 333 places using the remaining 444 digits, without repetition. Number of ways: 4P3=4×3×2=24^4P_3 = 4\times 3\times 2 = 244P3​=4×3×2=24

  5. Therefore, total number of such numbers is: 3×24=723\times 24 = 723×24=72

  6. Hence, the correct option is: 72\boxed{72}72​ which is option C.

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