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how many possible combinations of digits 1,2,3,4 are there?

stopi 43697 8
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  • #1 6467047
    stopi
    Level 28  
    Posts: 1691
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    I didn't know in which department to ask this question, so I put the programmers in the department, maybe they will know.

    Namely: how many possible combinations of numbers 1-2-3-4 are there, so that in each of these combinations none of them is repeated ..?

    Is it the so-called variation without repetition or n-word combination of a k-element set? I am not sure how to calculate it. Probability account I had a long time ago and I don't remember it exactly.

    It definitely doesn't fit the General Programming section, I'm moving to Hyde Park - krzychoocpp
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  • #2 6467057
    mr.Sławek
    Level 32  
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    24
  • #3 6467462
    stopi
    Level 28  
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    I wrote them all out and it turns out that it is :) Apparently it is a "k-element variation without repetition from an n-element set" defined by the formula

    n! / (nk)!

    in this case, a 4-word variation from a 4-element set.
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  • #4 6468765
    dg.1607
    VIP Meritorious for electroda.pl  
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    4*3*2*1 = 24
  • #5 6469028
    SeLuMar
    Level 14  
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    So for the sake of nomenclature, this is neither a variation nor a combination, but a permutation. The number P of possible combinations of n elements is given by the formula
    P = n!
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  • #7 6469269
    stopi
    Level 28  
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    SeLuMar notice that for n = k this my "variation" just gives n! Because the denominator has 0! = 1.
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  • #8 6469288
    Anonymous
    Anonymous  
  • #9 6469627
    stopi
    Level 28  
    Posts: 1691
    Help: 39
    Rate: 251
    Quote:
    how many possible combinations of numbers 1-2-3-4 are there, so that in each of these combinations none of them is repeated ..?


    Well, that's what I wrote at the beginning :)
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Topic summary

✨ The discussion revolves around calculating the number of possible combinations of the digits 1, 2, 3, and 4, ensuring that no digit is repeated in each combination. The consensus is that this scenario represents a permutation of 4 elements, which can be calculated using the formula P = n! (n factorial). For the digits 1-4, the total number of permutations is 4! = 24. The conversation also clarifies that if the task involved selecting a subset of these digits (e.g., combinations of 2 or 3 digits), it would be classified as a variation without repetition.
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