What is the probability that a positive divisor of $8748$ million is the product of exactly $20$ non-distinct primes?
To try and solve this question I split up $8748$ into $2^8 \cdot 5^6 \cdot 3^7 $ and came up with $4/(9 \cdot 8 \cdot 7)$ as an answer. where am I going wrong?
How did you get to the result $4 / (9\cdot 8\cdot 7)$?
$9\cdot 8\cdot 7$ is the number of all divisors of $N = 8748000000$, which you probably calculated yourself. Now, you must calculate how many of these divisors contain exactly $20$ non-distinct primes. This means answering the question:
In how many ways can I select $3$ numbers $a,b,c$ which add to $20$ and obey the fact that $0\leq a \leq 8,$ $0\leq b \leq 7,$ $0\leq c \leq 6$? How many ways can you do that? Can you list the $4$ you found?