Mobile phone operator zim" in my country is saying it has more than four million clients.

If a phone number is 9 digits long and always starts with a 6...

How is this possible?

For a number with two digit, if the first is 6, there are ten evens out.

6,1, ..........6,0. 1x10.

For a number with nine digit, the first is 6, so 10^8 possible evens.

There are 9 places to put digits in.

The first place always has a 6.

The second place can have any one of ten different digits using these: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

The third place can have any one of ten different digits using these: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

And so on . . .

 

So there are 10 possible digits for the second place. 10 for the third and so on . . . up to the 9th place

 

place . . . . . . = 2 . . 3 . . 4 . . 5 . . 6 . . 7 . . 8 . . 9

possible digits = 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 10^8

 

This is 10 000 000 = 10 million. They're telling it how it is

 

This is not a factorial type problem

If you need to know about this ask