Explanation

(c)
The number between 999 and 10000 are of four digit numbers.
The number of four digit numbers formed by digits 0, 2, 3, 6, 7, 8 is $\ ^{6}P_{4} = 360$
But here those numbers are also involved which being 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$= 360 - 60 = 300$