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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.

Aspiring Mathlete
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3 Answers3

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

Bhaskar Vashishth
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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.

David
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1) Chipola College must assign a new employee a 7 digit phone number, but the first three digits must be 526. How many different numbers can be assigned if:

a) There are no restrictions on the remaining four numbers?

b) the phone number cannot have any digits repeated?

c) the new employee is in the math dept and all math dept phone numbers end with 00?

unknown
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