0

For a data set of integers from 1 to 48, $$\binom{48}{4} = 194,580$$ distinct combinations.

And for another data set of 48 integers with 4 subsets of integers from 1 to 12 in each subset, a $$\binom{4\cdot 12}{4} = 194,580$$ as well, but the combinations are NOT distinct because the data set has repeated elements (integers 1 to 12 repeat 4 times).

Here is my question, what general method or formula can be used to count or determine distinct combinations for data set with repeated elements?

gt6989b
  • 54,422
user469843
  • 47
  • 1
    To be clear, you want to allow sets such as $1,1,2,3$ to be chosen, but that would be counted as the same as $1,3,2,1$? – kccu Feb 04 '19 at 18:37
  • What combinations are you counting? (Especially in your second example) – Jakob W. Feb 04 '19 at 18:59
  • Set (1,1,2,3) is not the same as (1,3,2,1). Example, set (1,1,2,3) can repeat x times, and set (1,3,2,1) can repeat y times, and I am interested x and y values and/or the number of unique sets from the total 194,580 combinations. – user469843 Feb 04 '19 at 19:31
  • If your sets (really ordered $4$-tuples) take the form $(x_1,x_2,x_3,x_4)$ where each $x_i$ is chosen from ${1,2,\dots,12}$, then there is only one way in which $(1,1,2,3)$ can be chosen. Similarly, there is only one way $(1,3,2,1)$ can be chosen. Perhaps you want to count the ways that $(1,1,2,3)$ can be rearranged? – kccu Feb 04 '19 at 19:49
  • Let me use a smaller data set to explain: A set of 9 integers consisting of 1,1,1,2,2,2,3,3,3 taking 3 items at a time has the following 84 combinations in 10 distinct groups (the x’s numbers in parentheses representing the number of repetitions -- note that 1x + 9x + 9x + 9x + 27x + 9x + 1x + 9x + 9x + 1x = 84 combinations).

    111(1x) 112(9x) 113(9x) 122(9x) 123(27x) 133(9x) 222(1x) 223(9x) 233(9x) 333(1x)

    What method or formula I can use to determine the n groups, and the x repetitions in each group?

     – user469843 Feb 05 '19 at 00:59

0 Answers0