Counting Techniques Question, Please Help!

Question

How many seven-digit telephone numbers can be made if the first three digits must be different

and the number cannot start with “0”?

I'm having some trouble with this question, can someone please explain how to do this? It would be very helpful, and for these type of questions how would you know if the elements can be repeated or not?

Answer 1

The first digit has 9 choices of 1-9, since 0 is not allowed. The second digit also has 9 choices since the first digit can't be reused. The third digit has 8 choices, since the first two can't be reused. The last seven digits each have choices 0-9 or 10 since digits can be reused.

N = 9×9x8x10x10×10×10 = 648 x 10⁴

Answer 2

There are a total of 7 spots in the phone number. As the condition states the first three digits must be different, and the number cannot start with 0, we can count each spot:

The first digit cannot be 0, so it can be any number from 1 through 9 which gives us 9 choices.

The second digit must be different from the first digit. It can be any digit from 0 to 9 except the first digit, so there are 9 choices left.

The third digit must be different from both the first and second digits, so there are 8 choices left.

Now, the last four digits have no restriction mentioned. Since the problem only says the first three digits must be different, the remaining digits can repeat. Each of the last four spots can be any digit from 0 to 9, so each has 10 choices. So, the total is:

9 x 9 x 8 x 10 x 10 x 10 x 10

= 6,480,000 numbers can be made

For these types of questions, assume digits can repeat unless the problem explicitly states that they cannot. Here, it specifically says the first three digits must be different, so only those first three are restricted. Since it does not say anything about the last four digits, repetition is allowed there.