Question

Need eight digit combination using 1,2,9,0,1,0,0,0.  How many possible combinations are there?  All numbers listed are part of the solution.  There must be two "1's", one "2", one "9", and four "0's".

Answer 1

If by combinations you mean as in a safe, it's permutations of those digits you want isn't it? If all the numbers were different, the answer would be 8 factorial (8!), which is 8*7*6*5*4*3*2*1 = 40,320. However, you've got some duplicate digits. So we have to reduce this number by the number of permutations we've "lost". That's 4! (24) for the four zeroes and 2! (2) for the two 1's. That leaves us with 840 when we divide by 48. The same answer would apply of course for any digits similarly duplicated. If asked by the police, I shall of course deny all knowledge!