Maybe not so easy, but interesting for sure. Some homework?
What the third parameter is for?
Just for fun, I will give a try.

It looks to me like you are counting...

mix("abc",2,"");
aa ab ac ba bb bc ca cb cc

could be replaced with
mix("012",2,"");
00 01 02 10 11 12 20 21 22


The characters you supply changes the base.
The number example above is base 3.

Like any number system the it would roll over. In this case from 22 to 00.



In binary would be
mix("01",2,"");
00 01 10 11 *roll over* 00 01 10 11 00 rinse and repeat.

Good catch NoahBuddy.
I always avoid recursion when it is not needed.
Here is the result, using simple iterations.
Code:

void setup()
{
  println(mix("abc", 2));
  println(mix("ab", 3));
}

String[] mix(String digits, int numberLength)
{
  int base = digits.length();
  int sequenceLength = int(pow(base, numberLength));
  println(sequenceLength);
  String[] result = new String[sequenceLength];
  String[] digitArray = digits.split("");
//  println(digitArray);
  for (int i = 0; i < sequenceLength; i++)
  {
	result[i] = MapDigits(i, numberLength, base, digitArray);
  }
  
  return result;
}

String MapDigits(int val, int numberLength, int base, String[] digitArray)
{
  String r = "";
  for (int i = 0; i < numberLength; i++)
  {
    int d = val % base;
    val = val / base;
    r = digitArray[d + 1] + r; // Small string, string concatenation OK here
  }
  return r;
}
 

yeah, recursive stuff is voodoo, i never feel in control of it. especially the stack usage.

that said, this wasn't that complex:

void mix(String alphabet, int n, String prefix) {
 if (n == 0) {
   // finished - prefix is entire string
   println(prefix);
   return;
 }
 for (int i = 0 ; i < alphabet.length() ; i++) {
   // add letter to what we have, reduce required length
   mix(alphabet, n - 1, prefix + alphabet.charAt(i));
 }
}

(if this is homework, don't use it if you don't understand it)