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Math Challenge II-A Combinatorics
It says that the answer is "We have 810000=24⋅34⋅54810000=24⋅34⋅54. Thus, we need to solve a+b+c=4a+b+c=4 a total of three times (once each for 2,3,52,3,5). Thus there are (4+3−14)3=3375(4+3−14)3=3375 different products.".
However, Shouldn't it be "5+3-1 choose 5" then quantity cubed, which is equal to the cube of 21 which is 9261? Because each exponent can have 5 different results, which are 0,1,2,3,4. There are five, not four numbers, between 0 to 4. Am I wrong or is the answer wrong?
Thanks!
$$810000=2^4\cdot 3^4 \cdot 5^4$$
As we split it into $3$ factors, each factor is of the form $2^? \cdot 3^?\cdot 5^?$, where the $?$ at the exponents may be from $0$ to $4$. However, we are not just counting one factor; the three factors should look like the following:
$$ 810000 = (2^a \cdot 3^m \cdot 5^x) (2^b\cdot 3^n\cdot 5^y) (2^c \cdot 3^p \cdot 5^z) $$
For the prime factor $2$, we need $a+b+c=4$. For $3$, we need $m+n+p=4$. For $5$, we need $x+y+z=4$. These are three Stars-and-Bars problems. The $4$ here is the total exponent itself, instead of the number of possible ways.
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