Say there are “n” numbers from which we select “p” numbers (p less than n) such that the selected “p” numbers are sorted. A selected number can be repeated. How can we calculate the number of combinations we can select? For example if we have a set of numbers say {1,2,3,4,5,6} (n=6) and we are to select 3 numbers from the set (p=3) which are sorted. So we can have {1,2,3}, {1,1,2}, {2,3,6}, {4,5,5}, {5,5,5} …….. Since all of these combinations are sorted they are valid. How can we find the number of such sorted combinations we can get?
What I mean from the word sorted, is that when we select p elements from a set of numbers of n elements, the selected p elements should be sorted.
Take this small example:
If the set is {1,2,3,4} (so n = 4) and we are to select 3 elements (p = 3), the number of ways we can select p elements (with replacement) will be 4*4*4=64. So the selections will have {1,1,1},{1,1,2},{1,1,3}{1,1,4},{1,2,1}.....{3,1,1}...{4,4,4}. But in these selections, not all are sorted. In this example, {1,2,1} and {3,1,1} are not sorted.
I want to get the number of sorted selections.
Thanks.
I fail to see how the sorting affects the result. For each possible combination with repetitions, there will be a corresponding sorted permutation.
Hence the question boils down to number of combinations of n elements taken p at a time with replacement. That is the straight forward formula, (n-1+p)C(p) = factorial(n-1+p) / (factorial(p) * factorial(n-1) )
Here is an explanation of the formula, and another one from wolfram.