Question

I need a function count_permutations() that returns the number of permutations of a given range. Assuming that the range is allowed to be modified, and starts at the first permutation, I could naively implement this as repeated calls to next_permutation() as below:

template<class Ret, class Iter> Ret count_permutations(Iter first, Iter last) {     Ret ret = 0;     do {         ++ret;     } while (next_permutation(first, last));     return ret; } 

Is there a faster way that doesn’t require iterating through all the permutations to find the answer? It could still assume that the input can be modified, and starts in the first permutation, but obviously if it is possible to implement without those assumtions it’d be great too.

Answer 1

The number of permutations for a range where all the elements are unique is n! where n is the length of the range.

If there are duplicate elements, you can use n!/(n_0!)…(n_m!) where n_0…n_m are the lengths of duplicate ranges.

So for example [1,2,3] has 3! = 6 permutations while [1,2,2] has 3!/2! = 3 permutations.

EDIT: A better example is [1,2,2,3,3,3] which has 6!/2!3! = 60 permutations.