Math Is Fun Forum

Danbee

Member

Registered: 2008-08-28

Posts: 21

Permutations and combinations

The telephone company has run out of seven-digit telephone numbers for an area code.To fix this problem, the company will introduce a new area code. Find the number of new seven -digit telephone numbers that can be generated for the new area code if both of the following conditions must be met:

CONDITION 1: The first digit cannot be 0 or 1
CONDITION 2: The first three digits cannot be the emergency number (911) or the number used for the information (411)

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Permutations and combinations

Without restrictions, there would be 10^3 = 1000 choices for the first three digits.
However, the first digit is not allowed to be 0 or 1, and those digits make up 100 choices each.
You're also not allowed 911 or 411, so in total there are 798 choices for the first three digits.

The last four are completely free, so the choices for that is simply 10^4 = 10000.

Therefore, the total number of possible telephone numbers is 7980000.

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Why did the vector cross the road?
It wanted to be normal.

musician_14

Member

Registered: 2008-11-01

Posts: 4

Re: Permutations and combinations

That is comfirmed.

There are 8x10x10 x 10x10x10x10 different telephone numbers generated using the first condition (no 0's or 1's in first digit)
8x10^6 = 8 000 000 .

Now for the second condition, we place the digits 911 and 411 as the first 3 digits and figure out how many ways this can be done.
For 911:  1x1x1 x 10x10x10x10.  This can be done in 10^4 = 10 000 ways.
For 411:  1x1x1 x 10x10x10x10.  This can be done in 10^4 = 10 000 ways.
Since we do not want this to happen, subtract this from the number we found in condition 1.

We obtain that

8 000 000 - 10 000 - 10 000 = 7 980 000

Thus, there are a total of 7 980 000 ways to generate the new seven digit numbers for the new area code.

Last edited by musician_14 (2008-11-01 13:22:24)