SOLUTION: if a five digit number is chosen at random, what is the probability that the product of the digits is 20?

Since 5 is a prime factor of 20, one of the digits must be 5

The remaining four digits must then have product 4

The only way four digits can have product 4 is for them 
to be 1,1,1,4 or 1,1,2,2 in some order.

The number of arrangements of 1,1,1,4 which is 4 things 
3 of which are indistinguishable is 4!/3! = 4

The number of arrangements of 1,1,2,2 which is 4 things 
2 pairs of which are indistinguishable is 4!/(2!2!) = 6

That's a total of 10 4-digit numbers which have product 
of digits 4.

Each of these 10 4-digit numbers have 5 places to insert 
a 5 to make it into a 5-digit number

[For example we can insert a 5 in 
the 4-digit number 2121 and make 
these 5 5-digit numbers 52121, 25121, 21521, 21251, 21215.)

Therefore there are 10*5 = 50 5-digit numbers with product
of digits 20.

The largest 5 digit number is 99999
The largest 4 digit number is 9999
So there are 9999 numbers with less than 4 digits.
So the number of 5-digit numbers is 99999-9999 = 90000

Therefore the desired probability is 50/90000 = 1/1800

Edwin