Reply With QuoteI think this is right for 537, as there would be 10 options for each subsequent number. Unless you have any rules on numbers not allowed etc (e.g like 5370000, 5371111) then you can minus off any like this.
Hey everyone, I have a question on a problem I have been assigned. Here is the problem:
A telephone number consists of seven digits, the first three representing the exchange. How many different telephone combinations are possible in the 537 exchange?
I came up with 10x10x10x10=10000
This seems too easy so I can't imagine I am right
What do you think? Thanks in advance.
Steve
I think this is right for 537, as there would be 10 options for each subsequent number. Unless you have any rules on numbers not allowed etc (e.g like 5370000, 5371111) then you can minus off any like this.
Ok, Stinker, thanks.
Sometimes, we treat the most simple things with a complicated way.Originally Posted by desmoface
you have 537**** well **** can take any number! that is from 0 to 9999; how many numbers are they? Well, 10000! simple as that ;-) (however the 10x10x10x10 notation is preferable in case you had ABCs and stuff :P)
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