How many 6 digit telephone numbers can be formed if each number starts with 35 and no digit appears more than once?
The first two places can only be filled by 3 and 5 respectively and there is only 1 way for doing this.
Given that no digit appears more than once. Hence we have 8 digits remaining
(0, 1, 2, 4, 6, 7, 8, 9)
So, the next 4 places can be filled with the remaining 8 digits in 8P4 ways.
Total number of ways = 8P4 =8×7×6×5=1680.