generate all possible combinations

[nikovlad]

Please help me generate all possible combinations of the given numbers without duplication of digits.

[k0st4din]

Hi welcome to the forum.
I ask - How do you combine the numbers on how many possible. I play the lottery or something else?
Your question is sort of - I am looking for something in infinity. Be particular to try to help.
Merry knees and Happy New Year

[Sisyphus]

Are these combinations like 3L, 8R, 1L?

[Sisyphus]

Hi,

If anything, your question has proven that I am a lousy mathematician.

I started out with a number lock. 3 number wheels, each with numbers from 0 to 9, enables 1000 combinations from 000 to 999. 1000 = 10^3, but the number of wheels isn't significant and this formula is accidental. The correct formula is 10 * 10 * 10 where each 10 represents the number of choices. Accordingly, if my lock had one wheel each of 4, 6 and 8 numbers then the number of choices would be 4 * 6 * 8. So, that should help you crack the safe: 3 wheels of 20 numbers each plus Right and Left for each should give you 20 * 20 * 20 * 2 combinations. That is why they put time locks on such safes. You can't just get through all the combinations before the damn thing disables the combination lock.

Then I took the numbers 1 to 3 and arranged them into possible sequences:
1-2-3, 1-3-2
2-1-3, 2-3-1
3-1-2, 3-2-1
I think I got them all. 6 is the total, but by which formula do I arrive at 6?
Too difficult. So, I tried something easier:
1-2
2-1
The formula seems to be: [Top number] * [Top number -1].
That gives me 2 * 1 = 2 or, in the first series, 3 * 2 = 6.

That is encouraging. Try another one:
1-2-3-4, 1-2-4-3
1-3-2-4, 1-3-4-2
1-4-2-3, 1-4-3-2
Observe that the last two numbers are always the same, just juxtapositioned, in the sense that they are the only two numbers not used by the rest of the string. Regardless of how many positions I have, the last two must always be the two I didn't use before. So I argue that the reason why I have 6 results all starting with 1 is because the remaining digits (2, 3 and 4) have has many choices as the 1-2-3 example. Therefore, the formula for calculating the number of combinations of a series of numbers from 1 to 4 should be 4 * 3 * 2 * 1 = 24. Any one wants to contradict me that the number of choices for the numbers from 0 to 5 = 6 * 5 * 4 * 3 * 2 * 1?
There are 52 cards in a Black Jack deck.

Of course, these are precise sequences. The number of combinations, if sequence is of no importance, is only as many as the number itself indicates. That seems to work with the numbers lock. The series represents a single wheel with 4 numbers on it, therefore the formula is 1 * 4. Correct.

So, did I answer the question? How many possible combinations are there? If sequence is of no importance it is the number of choices multiplied by the number of choices, multiplied by the number of choices [in perpetuum].
If that answer doesn't remove the last bit of understanding you had before I should try again.
Have a great day! It's the last one you get this year.