Number of different possible permutations of a telephone number

Question

A telephone number consists of $10$ digits, all from $0$ to $9$. The first digit is $0$. The remaining digits can be any number ranging from $0$ to $9$. How many possible telephone numbers are there?

My try:

I first said that since the first digit is $0$, we only need to look at the remaining $9$ digits and see how many different ways they can be arranged. If the $9$ remaining numbers can be from $0$ to $9$. Then The possible arrangements are:

$$ 1 \times (9! \times 9!) \times 9 $$

The last $9$ refers to the remaining $9$ digits. Can someone please elaborate on my answer? I'm not sure if I am correct.

Answer 1

No. First is fixed, so for next 9 places you have ten choices each. so total are $1 \times ( 10 \times 10 \times 10 ... \times 10)=10^9$

Answer 2

Let's first look at $3$ digit phone numbers to simplify. If the first digit is "locked" at $0$ only, we then have $10$ choices for the $2$nd digit and $10$ choices for the $3$rd digit, thus giving us $10$ x $10$ = $10^2$ = $100$ possible phone numbers. Examples would be $000, 001, 012... 010, 011, 012...090, 091, 092...099$

For $10$ digit phone numbers with the first digit "locked" at $0$ it is similar except it would be $10^9$ = $1,000,000,000$ = $1$ billion.