Question

How many 6-digit telephone numbers can be constructed with digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 if each number starts with 35 and no digit appears more than once?

Answer 1

Total number of digits = 10
The first 2 digits of telephone is 35 and no digit appears more than once.
$\therefore$ Total number of remaining digits = 10-2 = 8
And, Total number of remaining digits of telephone number = 6-2 = 4
Required number of telephone numbers
= $8_{P_4}$
= $\frac{8!}{(8-4)!}$
= $\frac{8!}{4!}$
= $\frac{8\times 7\times 6\times 5\times 4!}{4!}$
=1680