If a campus telephone extension has four digits, how many different extensions are there with no repeated digits:

1. If the first digit cannot be 0?

We can pick the first digit out of 9 digits (since we cannot use 0) and the three remaining ones also from 9 digits (this time 0 is included but the digit the we picked for the first position cannot be used so we still have 9 choices):

\[9 \cdot P(9, 3) = 9 \cdot 9 \cdot 8 \cdot 7 = 4536\]

2. If the first digit cannot be 0 and the second cannot be 1?

We sum up two cases:

1. The first digit is 1, then we have 9 choices for the second one, 8 choices for the third one and 7 choices for the fourth one: \(1 \cdot 9 \cdot 8 \cdot 7 = 504\).
2. The first digit is between 2 and 9 (8 choices), then we have 8 choices for the second one (cannot be 1 or the same as the first digit), 8 choices for the third one (cannot be the same as the first two digits) and 7 choices for the fourth one: \(8 \cdot 8 \cdot 8 \cdot 7 = 3584\)

for the total of \(4088\)