## March 2, 2010

### Phone number permutation

If you were to dial any 7 digits on a telephone in random order,
what is the probability that you will dial your own phone number?

Assume that your telephone number is of 7-digits.

Tiger Woods said...

1 in (10^7)

Anonymous said...

1 in 9x10^6

Anonymous said...

1 in 9000000 cuz its observation over the total frequency and your phone number is one number and the total possible numbers is 900 0000 (100 0000 to 999 9999)

Jakob said...

1/(10^7), since there are 10 numbers (counting 0), and there are 7 digits in the number.

Secret SquÃ¯rrel said...

It is 1 in 9x10^6 (1 in 9,000,000) because no-one's local phone number begins with zero. Zeroes are reserved for signalling the exchange that you are making a special type of call; eg emergency (000), out-of-area (0n nnn nnnn), mobile/cell (0nnn nnn nnn), etc.

Anonymous said...

1/(7x10^N).1 represents your phone number,7 represents the number of digits of your phone number, 10 represents the numbers that can be dial in a telephone and the N represents that how many times do you dial before you could possibly dial your phone number.....

ulquiorra said...
This comment has been removed by the author.
Anonymous said...

To start off , let us tackle numbers with 2 digit only.

1st digit 2nd digit
can be 1-9 can be 0-9

because nos don't start with 0 for that will make the number a one digit entity.So toally 9*10 = 90 nos

for 3 digit nos

1-9 0-9 0-9

9*10*10 = 900

Similarly for 7 digit no,it will be

9*10*10*10*10*10*10*10
= 9 * 10^6

= 9* 10 ^ (no of digits -1 )
Hence our probability is

1 / 9*10^(7-1) = 1/ 9*10^6

### ABHIRAM ### said...

since 1st digit can not be zero
total no of ways=9*10*10*10*10*10*10
=9*10^6
probability=1/9*10^6

Anonymous said...

one mill of a chance
10^7