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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?

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Solution

Total number of digits = 10
The first 2 digits of telephone is 35 and no digit appears more than once.
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
= 8P4
= 8!(84)!
= 8!4!
= 8×7×6×5×4!4!
=1680


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