Heres a math problem that makes you want to die :) solve it please?

Question

How many ten digit phone numbers are possible if the first digit cant be zero, the first three digits cant be 800 or 900, and the number must end in 0000.

p.s. answer is NOT 567,000

Answer 1

The first 3 digits cannot be 800, or 900, or all the numbers from 000 to 099. That leaves a total of 1000 - 102 = 898 possibilities left. The next 3 digits can be anything from 000 to 999, or 1000 possibilities. The last 4 digits must be 0000, or just one possibility. The answer is the product: 898 x 1000 x 1, or 898,000.

Answer 2

You mean the phone number must NOT end in 0000, correct?

It's not really that hard. There are 898 possibillilties for the first three digits (eliminate the 100 that start with a zero, along with 800 and 900). The next three can be anything, so there are 1,000 possibilties. The last four can be anything except 0000, so there are 9,999 possibilities.

To get the answer calculate 898 * 1000 * 9999.

Answer 3

Ok- I'll give you the formula to figure it out.

We have ten digits, but the last four are zeros- so we ignore them!

The first digit cannot be 0, 8 or 9- so we have 1,2,3,4,5,6, and 7 left to work with- 7 possibilities!
so we write for the first digit: 7C1

The second and third can be anything from 1 to 9, just not 0. W have 9 possibilities! Because order matters, we use the P function.
so we write 9P2

Now we have two digits left- and they can be anything from 0 to 9. So we have 10 possible for this!
Because order matters, we use the P function.
so we write 10P2

So all in all, you multiply through:
7C1*9P2*10P2

Figure it out from there... my calculator has gone missing... :)

Answer 4

898*1000 = 898000

Answer 5

heres the answer 452546740800000 no not really the answer is 898,764

Answer 6

567,001 ha