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BITSAT 2021
Exam
BITSAT
Year
2021
Subject
Mathematics
Chapter
Permutation and Combination

BITSAT 2021 · Mathematics · Permutation and Combination

How many numbers lying between 999 and 10000 can be formed with the...

How many numbers lying between 999 and 10000 can be formed with the help of the digits 0,2,3,6,7,80,2,3,6,7,8, when the digits are not repeated?

Options

  1. A100
  2. B200
  3. C300
  4. D400

Correct answer

C. 300

Step-by-step solution

The numbers between 999 and 10000 are all 4-digit numbers. The number of 4 - digit numbers formed by digits 0,2,3,6,7,80,2,3,6,7,8 is 6P4=360{ }^{6} \mathrm{P}_{4}=360. But here those numbers are also involved which begin from 0. So, we take those numbers as three-digit numbers. Taking initial digit 0 , the number of ways to fill remaining 3 places from five digits 2,3,6,7,82,3,6,7,8 are 5P3=60{ }^{5} \mathrm{P}_{3}=60 So, the required numbers =360−60=300=360-60=300.
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