Question

Suppose account numbers for an internet service provider consist of alpha-numeric characters. The first character is a digit (0 through 9). The next 2 characters are capital letters (A through Z) and the last four characters are digits. How many account numbers are possible if no digit appears more than once? Your Answer:

Answer 1

Answer:

20,442,240 account numbers are possible.

Step-by-step explanation:

The first character has 10 possible values.

The next 2 has 26 * 26 = 676 possible combinations.

The last 4 has 9*8*7*6 = 3024 possible combinations.

So the answer is 10 * 676 * 3024

= 20,442,240.

Answer 2

There exist 20,442,240 possible combinations if no digit appears more than once.

How to estimate the account numbers that are possible if no digit appears more than once?

The first character contains 10 possible values.

The next 2 contains 26 $*$ 26 = 676 possible combinations.

The last 4 have 9 $*$ 8 $*$ 7 $*$ 6 = 3024 possible combinations.

The has 10 $*$ 676 $*$ 3024 = 20,442,240 possible combinations.

Therefore, there are 20,442,240 possible combinations if no digit appears more than once.

To learn more about combinations

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