Question

I have a language consisting of the words having exactly two “1” and three “0”. How can I efficiently enumerate the finite set of all words of this language?

Answer 1

Easy, write the number 11100, calculate the number of permutation of this value = n! = 5!,
divide by the number of permutation of the 3 1’s = 3! and the number of permutation of 0’s = 2! => 5! / (2! * 3!) = 120 / (6 * 2) = 10

11100
11010
11001
10110
10101
10011
01110
01101
01011
00111

Now if you need the actual values, for an arbitrary language, you have no other choice but to use a backtracking algorithm.

For this particular case, you can easily build a simple algorithm generating this language:
Here’s an example using python

def GenerateLanguage(nZeros, nOnes):
    if nZeros + nOnes == 0:
         return ['']
    res = [] # Resulting list, initialize with 1 empty string
    if nOnes > 0: # If we have 1's left, build all the strings that starts with a 1
         for l in GenerateLanguage(nZeros, nOnes - 1):
              res.append('1' + l)
    if nZeros > 0: # If we have 0's left, build all the strings that starts with a 0
         for l in GenerateLanguage(nZeros - 1, nOnes):
              res.append('0' + l)
    return res