BITSAT 2021 Paper

The numbers between 999 and 10000 are all 4-digit numbers. The number of 4 - digit numbers formed by digits $0,2,3,6,7,8$ is ${}^{6}P_{4}=360$. 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 $2,3,6,7,8$ are ${}^{5}P_{3}=60$ So, the required numbers $=360-60=300$.