Calculating pandigital number combinations

Damienj

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Jul 5, 2015
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Now, I know normally pandigital numbers do not take 0 into account, but in my case it must be taken into account. Do we have ten digits to work with {0,1,2,3,4,5,6,7,8,9} . My question is, how does one calculate the amount of possible combinations with these numbers such that each number cannot be repeated in a particular combination. Examples of such combinations include: 0195278436 or 4765213809. I'm thinking that the way to work this out is 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 which equals 3,628,800 possible combinations. I am aware that if the numbers could repeat it would be 10x10 ten times over which would give 10,000,000,000 combinations but they can't repeat so taking one away from the next number being multiplied seemed like the right way to do it. Does anyone know how to figure this out properly? Or have I done so already? Sorry if this is in the wrong section but I'm new to the forum and I haven't seen a sequences and series section after a quick glance.
 
...My question is, how does one calculate the amount of possible combinations with these numbers such that each number cannot be repeated in a particular combination. Examples of such combinations include: 0195278436 or 4765213809. I'm thinking that the way to work this out is 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 which equals 3,628,800 possible combinations...

I see no reason why your intuition would lead you astray. You're essentially saying "Okay, for the first place, I can choose any of the ten digits. Then for the second place, there's only nine digits remaining to choose from, etc." So, the number of pandigital numbers (if we include the digit 0) seems to be 10! or 3,628,800.
 
I see no reason why your intuition would lead you astray. You're essentially saying "Okay, for the first place, I can choose any of the ten digits. Then for the second place, there's only nine digits remaining to choose from, etc." So, the number of pandigital numbers (if we include the digit 0) seems to be 10! or 3,628,800.
Ahh I feel so silly now, I had the answer all along! Thanks for the clarification, it's much appreciated
 
Now, I know normally pandigital numbers do not take 0 into account, but in my case it must be taken into account. Do we have ten digits to work with {0,1,2,3,4,5,6,7,8,9} . My question is, how does one calculate the amount of possible combinations with these numbers such that each number cannot be repeated in a particular combination. Examples of such combinations include: 0195278436 or 4765213809. I'm thinking that the way to work this out is 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 which equals 3,628,800
Surely you must know about factorials?
 
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