How many different 7 digit numbers can be formed using the digits 1, 3, 0, 3,…

Final answer: 360

Steps to compute:

• Count all permutations of the seven digits accounting for repeated digits. The multiset is {1, three 3s, two 5s, 0}, so total permutations = 7! / (3! · 2!) = 5040 / 12 = 420.

• Exclude arrangements that start with 0 (these are not 7-digit numbers). Fix 0 in the first position and permute the remaining 6 digits: 6! / (3! · 2!) = 720 / 12 = 60.

• Valid 7-digit numbers = 420 − 60 = 360.

Explanation: Use the formula for permutations of multiset elements to handle repeated digits, then subtract arrangements with a leading zero because those do not produce valid 7-digit numbers.