The permutations of a set are the ways of arranging its elements.

For example, there are 24 permutations of a set of 4 elements:

[0, 1, 2, 3]
[0, 1, 3, 2]
[0, 2, 1, 3]
[0, 2, 3, 1]
[0, 3, 1, 2]
[0, 3, 2, 1]
[1, 0, 2, 3]
[1, 0, 3, 2]
[1, 2, 0, 3]
[1, 2, 3, 0]
[1, 3, 0, 2]
[1, 3, 2, 0]
[2, 0, 1, 3]
[2, 0, 3, 1]
[2, 1, 0, 3]
[2, 1, 3, 0]
[2, 3, 0, 1]
[2, 3, 1, 0]
[3, 0, 1, 2]
[3, 0, 2, 1]
[3, 1, 0, 2]
[3, 1, 2, 0]
[3, 2, 0, 1]
[3, 2, 1, 0]

To find the next permutation of an array ar:

1. Find the highest index, i₁ such that ar[i₁] is the first of a pair of elements in ascending order.
   If there isn’t one, the sequence is the highest permutation, so reverse the whole thing to begin again.

2. Find the highest index i₂, such that i₂ > i₁ and ar[i₂] > ar[i₁].
3. Swap ar[i₁] and ar[i₂].
4. The elements from ar[i₁ + 1] to the end are now in descending order (a later permutation), so reverse them.

static void swap(unsigned int *ar, unsigned int first, unsigned int second)
{
	unsigned int temp = ar[first];
	ar[first] = ar[second];
	ar[second] = temp;
}

static void reverse(unsigned int *ar, size_t len)
{
	unsigned int i, j;

	for (i = 0, j = len - 1; i < j; i++, j--) {
		swap(ar, i, j);
	}
}

unsigned int next_permutation(unsigned int *ar, size_t len)
{
	unsigned int i1, i2;
	unsigned int result = 0;
	
	/* Find the rightmost element that is the first in a pair in ascending order */
	for (i1 = len - 2, i2 = len - 1; ar[i2] <= ar[i1] && i1 != 0; i1--, i2--);
	if (ar[i2] <= ar[i1]) {
		/* If not found, array is highest permutation */
		reverse(ar, len);
	}
	else {
		/* Find the rightmost element to the right of i1 that is greater than ar[i1] */
		for (i2 = len - 1; i2 > i1 && ar[i2] <= ar[i1]; i2--);
		/* Swap it with the first one */
		swap(ar, i1, i2);
		/* Reverse the remainder */
		reverse(ar + i1 + 1, len - i1 - 1);
		result = 1;
	}
	return result;
}