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Math Challenge II-A Combinatorics

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Re: Math Challenge II-A Combinatorics
by Dr. Kevin Wang - Sunday, September 12, 2021, 3:15 PM

$$810000=2^4\cdot 3^4 \cdot 5^4$$

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.