SOLUTION: Think of a two-digit number. What is the probability that it has different digits?

Two ways to do it:

First way:

There are 99 numbers from 1 through 99. But the first 9 are 
1-digit numbers so the number of 2-digit numbers is 99-9=90

The ones that have the digits the same are the 9 multiples of 11,
ie., 11,22,33,...99

So there are are 90-9 = 81 2-digit numbers that have different 
digits.

Answer: 81 out of 90 = 81/90 = 9/10

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Second way:

First we find the number of two digit numbers with different digits:

Choose the first digit any of 9 ways (CANNOT choose 0).
Choose the second digit any of the 9 remaining (CAN choose 0).

So there are 9∙9 = 81 two-digit numbers with different digits.

That's the numerator of the probability.

Second, we find the number of two digit numbers:

Choose the first digit any of 9 ways (CANNOT choose 0).
Choose the second digit any of the 10 digits. (CAN choose 0).

So there are 9∙10 = 90 two-digit numbers.

That's the denominator of the probability.

Answer: 81/90 = 9/10  

Edwin