In how many ways I can express any integer in the form $N=a\times b$ I have any integer N which expressed in their primes, so
$N = p1 \cdot p1\cdot p2 \cdot p2 \cdot p2 \cdot p2 \cdot p3 \cdot p3$
where $p1, p2, p3$ are prime factors of N,
say $8100=2\cdot 2\cdot 3\cdot 3\cdot 3\cdot 3 \cdot 5\cdot 5$
I want to know in how many ways I can express the integer N in exactly their products of 2 factors, like that $\rightarrow N = a1\times a2$
,so from above example,  this can be done like this.
$
8100 = (2.2)(3.3.3.3.5.5) = 4\times 2025$
$8100 = (2.3.3)(2.3.3.5.5) = 18\times 450$
$8100 = (2.2.3.3)(3.3.5.5) = 36\times 225$
$8100 = (2.3.5.5)(2.3.3.3) = 150\times 54$
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I think you get the point;-)
I know little about permutation and combination, I think this can be solve by using combination.
I think I have to choose $\frac{1}{2\times 4 \times 2}(^8C_1 + ^8C_2+ ^8C_3+ ^8C_4)$ways(its wrong), but I know there are some repetitive number of prime (like 3 occur four times), I don't know how to do it, I just can't figure it out, I want your help.
I want to know the more generalise formula for any integer N.
 A: Start with prime factorisation but use a compact exponent form (or count the number of times a given prime occurs).
Let's say you're dealing with $N = p^aq^br^c$ where $p, q, r$ are primes.
The number of divisors of $N$ can easily be shown to be $m= (a+1)(b+1)(c+1)$. This formula can be extended to any number of prime factors.
If you're looking to express $N = xy$ with $xy$ being indistinguishable from $yx$, then the answer is $\lceil \frac m2 \rceil$. The ceiling function is necessary to take care of the case when the number of divisors is odd, which implies that $N$ is a perfect square.
If, on the other hand, you treat $xy$ and $yx$ as two distinct things (except when $x=y$ which occurs once with square $N$), then the answer is simply $m$.
If you're imposing additional conditions like the exclusion of $1$ and $N$, you can easily adjust the above results yourself.
A: Let us make a few assumptions as they are not mentioned in the question. For any number $a$ , we are not going to consider $a*1 \space and \space  1*a$. Also if $a=p*q$ is considered $a=q*p$ won't be considered. Suppose $a$ has $x$ no. of factors, for reach factor $c$ we can get a pair $(c,a/c)$, and if a is not a perfect square, then $(c,a/c)$ and $(a/c,c)$ are distinct. So total no of such distinct pairs will be $x/2$. If $a$ is square, then when $c=\sqrt(a)$, $(c,a/c)$ and $(a/c,c)$ are the same. So all other $c$ which are $x-1$ in numbers, gives two pairs. By same logic as above for those $x-1$ factors we get $(x-1)/2$ pairs. Adding the case we left out about both the entry in the pair being same, we get $(x-1)/2 +1 =(x+1)/2$. To accompany both the cases we can use the ceiling function! Figure out how!
Now the question boils down to finding out $x$,no of distinct factors excluding 1 and the number itself. That you can work out easily.
