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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?

user161367
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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?

5xum
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  • 8+5+7=20 8+6+6=20 are those the only 2? – user161367 Jul 31 '14 at 12:03
  • 7+7+6 is another – 5xum Jul 31 '14 at 12:15
  • And that's not all. – Gerry Myerson Jul 31 '14 at 12:50
  • @GerryMyerson It is as far as I know... If the first number is $\leq 6$, then the sum of the other two must be at leat $14$ which is impossible. If the first number is $7$, then the other two can only sum to $13$ if they are $6$ and $7$, if the first is $8$, then we have $8,7,5$ and $8,6,6$ and nothing else.. – 5xum Jul 31 '14 at 13:31
  • Never mind, I just realized that 1) you wrote $0\le a\le6$ when you meant $0\le c\le6$, and OP wrote $8+5+7$ when what was meant was $8+7+5$. – Gerry Myerson Jul 31 '14 at 13:35