numbers

[Yinzer]

What is the shortcut to figure our the total possible combinations of numbers given the amount of digits. More specifically my husband and I wondered how many possible phone numbers could arise out of seven digits (all within one area code). But I'd like to know the equation (or whatever) just for future reference.

[kags]

I think (and am happy to be corrected!) that if it is 7 digits the sum would be 7x6x5x4x3x2x1=number of combinations.

[kags]

I think (and am happy to be corrected!) that if it is 7 digits the sum would be 7x6x5x4x3x2x1=number of combinations.

[kags]

sorry about the echo!

[kags]

If there are 2 or more digits the same I think you just omit the multipliers equvalent or less than the number of duplicated digits - ie if there were 3 numbers the same in the 7 it would be 7x6x5x4. If there is more than one number duplicated then I have absolutely no idea - Bernardo, we need you!!!

[jenstar]

A simpler puzzle than you think: It's just the counting numbers! With seven digits there's 10 million combinations if you allow leading zeros (this includes 0000000) and 9 million (9999999-999999=9000000) if you don't.

[Yinzer]

What about letters, for example using A B and C in 4 (ie ABCC, BBCC etc). I remember learning SOMEthing about it back in little school but I wish I knew what it was!

[kags]

Same for letters - if there are 4 different letters then there are 24 combinations ie 4x3x2x1. If one is used twice then omit the 2 and the 1 and you get 4x3.

[New Forester]

As Jenstar says, with numerals it's the counting numbers. Ordinary numerals are of course in base 10. The number of possibilities for a 7 digit number (including leading zeros) is 10 to the power of 7. With letters, just think of them as numbers in base 26, and do the same thing. Count the number of letters in your "number". Raise 26 to this power. So the number of possible combinations for a three "digit" "number" would run from AAA to ZZZ. It would be 26 to the power of three, or 17,576 (there are no leading zeros to miss out of course). For other bases, you'd do the same. For example, I had a car with a radio code which worked off the pre-set channel buttons. There were five of these, so that was four digits in base five, or only 625 possible numbers. It didn't take me long to find by trial and error. Likewise, if you were using letters, but were not allowed some (as in number plates), you'd use the number of allowable ones as your base. The example above assumes you count to ZZZ starting from AAA. However, if you were counting like spreadsheet columns it would go: "A, B ... Z, AA, AB .. ZZ, AAA, AAB ... ZZZ". This is more complicated, as you are using imaginary leading zeros (00A, 00B etc) but not then using zero as a digit. I think this would be 26 +(26**2)+(26**3).

[New Forester]

Kags I'd assumed that Yinzer was using the whole alphabet. If so, you'd use up your 24 just getting from AAAA to AAAY. With just three letters allowed, it's 3 to the power of 4, or 81.

[Hammer]

I agree with New Forester, that for phone numbers it would be 10 to the power 7 for a seven digit number. The previous posts (7x6x5x4x...) only apply if numbers cannot be repeated, but they can for phone numbers. One other thing to remeber is that phone numbers cannot start with a zero, so there would be 9x10x10x10x10x10x10 = 9,000,000 combinations. There would also be a couple of others you couldn't start with (eg 999) so there would be a few less than 9 million possibilities.

[kags]

yup - my system only works to find out the combinations of 7 known digits or letters - and even then it doesn't work if more than one digit is duplicated. My head hurts!

[bernardo]

Why all the confusion? What's wrong with just doing 10 to the power of however many digits it is? It seems obvious to me that there are 10 combinations from 0 to 9, 100 from 00 to 99, 1000 from 000 to 999 [etc] and 10,000,000 from 0000000 to 9999999.

[New Forester]

As Hammer says, some numbers cannot be allowed at the start of phone numbers -- not just 999, but also special "operator" numbers like 100, 151, 152 etc. Also, many exchanges use only certain numbers in certain areas (I suppose derived originally from the local dialling codes). Ringwood, for example, is 01425, then 47[xxxx] or 48[xxxx], while New Milton is also 01425, but then 81[xxxx]. So the number of numbers available will be very much less than the theoretical maximum -- I guess perhaps 50,000 or so.

[New Forester]

It's easy for you, bernardo, you can do sums...

[Gef]

"More specifically my husband and I wondered how many possible phone numbers could arise out of seven digits (all within one area code)." The answer to this part of the question is very simple. For example in Glsagow where the area code is 0141 then xxx xxxx can have a maximum of 10 million combinations.

[Upforit]

I might be wrong here, but assuming the formula 10 to the power of seven is the correct formula:- Isn't 10 to the power of 7 the same as saying 10 with seven zero's after it, whic is 100 000 000 or one hundred million?