In how many possible ways can we write $3240$ as a product of $3$ positive integers $a$, $b$ and $c$? In how many possible ways can we write $3240$ as a product of $3$ positive integers $a$, $b$ and $c$? 

This is the question where I've been stuck. The answer is $450$, but I don't know why. I've tried taking out the number of factors, then applying the combination formula in different ways.
$$3240=2^3\cdot3^4\cdot5^1$$



The factor $2$ can be split among $3$ divisors in $10$ different ways:


$0,0,3$
$0,1,2$
$0,2,1$
$0,3,0$
$1,0,2$
$1,1,1$
$1,2,0$
$2,0,1$
$2,1,0$
$3,0,0$




The factor $3$ can be split among $3$ divisors in $15$ different ways:


$0,0,4$
$0,1,3$
$0,2,2$
$0,3,1$
$0,4,0$
$1,0,3$
$1,1,2$
$1,2,1$
$1,3,0$
$2,0,2$
$2,1,1$
$2,2,0$
$3,0,1$
$3,1,0$
$4,0,0$




The factor $5$ can be split among $3$ divisors in $3$ different ways:


$0,0,1$
$0,1,0$
$1,0,0$




Hence the total number of ways to write it as a product of $3$ divisors is $10\cdot15\cdot3=450$.
3420 can be written as 2^3 . 3^4 . 5^1

We have 3 places _ _ _ or lets denote them as x y z. 

We have to distribute 3 '2's, 6 '3's and 1 '5' into 3 places, where '2','3','5' are objects.

Now, How many ways to distribute 3 '2's ? You observe that no matter how you distribute '2's the total number of '2's will always be 3.

Lets say you distribute the '2's such that x gets 1 '2', y gets 1 '2', z gets 1 '2' like so ----- 2|2|2

The bars (|) separate what x, y, z gets like so ----------------------------------------------- x|y|z

Again you can distribute such that x gets no '2', y gets 2 '2's, z gets 1 '2' like ----------- |2.2|2

-------------------------------------------------------------------------------------------------- x| y |z

Again you can distribute such that x gets 3 '2's, y gets no '2', z gets no '2' like --------- 2.2.2||

------------------------------------------------------------------------------------------------------- x|y|z

Therefore you see this is an arrangement problem of 5 items with 3 similar items (3 '2's) and 2 similar items (2 bars(|))

Therefore no. of ways to distribute 3 '2's to 3 places will be 5!/(3! . 2!) = 10 ways

Similarly for distributing 4 '3's to 3 places will be arranging ||3.3.3.3 ; which is 6!/(4! . 2!) = 15 ways.

No. of was to distribute 1 '5' to 3 places is 3 ways

Therefore the total no. of ways is 10.15.3 = 450 ways.