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Math Challenge II-A Combinatorics

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Math Challenge II-A Combinatorics
by Zeyin Wu - Saturday, September 11, 2021, 8:02 PM
Question 3.26: How many different ways are there to represent 810000 810000 as the product of 3 3 factors if we consider products that differ in the order of factors to be different?

It says that the answer is "We have 810000=243454810000=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+314)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?


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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.