20 dials, each w/ 60 settings: find total num. combinations

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Kprime

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Oct 18, 2008
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UG, I hope this is the right forum. Ok I am not a math anything but I need to know how to arrive at an answer for what I think is a very simpleton problem.

I have an instrument with 20 dials. Each dial can be set to one of 60 different positions. How many total combinations can there be?

I actually need to know the formula. I can use that in an excel spreadsheet and compute new answers for other equipment with fewer dials.

Thank you kind folks for donating your time on this. Many patients of mine will benefit from your answer. I will be sure to cite your work in my writing.

Thank you, Thank you, Thank you. :D
 
Re: 20 variables

Hmm, I watched some of the math videos. I think the formula is 20^60.

Interesting, thats not the same as 60^20. So 20 dials with 60 settings each has fewer possibilities than 60 dials with 20 settings each. Ok. Can you confirm this conclusion?
 
Re: 20 variables

Kprime said:
Interesting, thats not the same as 60^20. So 20 dials with 60 settings each has fewer possibilities than 60 dials with 20 settings each. Ok. Can you confirm this conclusion?
Click to expand...
Why do you think that \(\displaystyle 20^{60} < 60^{20}\)?
 
Re: 20 variables

ooops, thats backwards, isn't it. I put those formulas into excel and the 20^60 is 1152921504606850000000000000000000000000000000000000000000000000000000000000000

the 60^20 is 365,615,844,006,298,000,000,000,000,000,000,000

so, yah, I got that backwards. :roll: In any event, I have to pay attention to the order because they produce different results. But is the approach correct? How do you state this problem? Variables?
 
Re: 20 variables

Suppose your instrument only had 3 dials each of which could be set to 10 positions.
You have 3 blanks representing the 3 dials.
__ __ __

How many positions go into the first blank?

10 __ __

How many positions go into the second blank?

10 10 ___

How many positions go into the third blank?

10 10 10

Answer --- 10[sup:szx19zga]3[/sup:szx19zga] = 1000

Does 3[sup:szx19zga]10[/sup:szx19zga] = 1000?
 
Re: 20 variables

No it doesn't, but 10^3 does. Not sure if that holds true with all examples. I can't proof it, but it could be right.
 
Re: 20 variables

hmmm, that seems I should go with 60 (positions) to the 20 dials 60^20.


curious what if the 3 dials had different values for the number of postions?

10 5 10
 
Re: 20 variables

Thank you very much. You have answered my questions for now and I think I have the right numbers to put into a paper I am writing. I appreciate your time and efforts. :!:
 
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