We need to find how many 7 digit codes are possible where each digit is 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 and the product of all the digits is 10000.
Do the prime factorization for 10000 to get = 2^4*5^4
We now know that 2,4,8,5,1 and 5 must be used 4 times
10000/5^4 = 16
I need 3 digits that multiply to 16
8*2*1
4*4*1
4*2*2
Therefore....
7!/4! =210
7!/(4!2!) = 105
7!/(4!2!) = 105
And we add these together to get 210+105+105 = 420 ways
Hope this helped!
Melody Jan 19, 2020