Number of ways of splitting $2310$ as product of three factors Find Number of ways of splitting $2310$ as product of three factors.
My Try:
$$N=2310=3 \times 7 \times 2 \times 5 \times 11$$ 
$1.$ if two of the factors are ones ten trivially it is one way.
$2.$ if exactly one of the factors in one  then number of ways is $\frac{(1+1)(1+1)(1+1)(1+1)(1+1)-2}{2}=15$ ways
Can i have any clue if None of the factors is one

We can choose prime factors for each of three factors.  I am assuming the order of the factors does not matter.


*

*All five primes are in one factor, leaving $0$ primes for each of the other two factors.  Note that this corresponds to the factorization $1 \cdot 1 \cdot 2310$.
$$\binom{5}{5}$$

*Four primes are in one factor and the remaining prime is in another factor.  In this case, one of the factors is $1$, one is prime, and the other is composite.
$$\binom{5}{4}\binom{1}{1}$$

*Three primes are in one factor and both of the remaining primes are in another factor.  In this case, one of the factors is $1$ and the other two are composite.
$$\binom{5}{3}\binom{2}{2}$$ 

*Three primes are in one factor and each of the remaining factors is prime.
$$\binom{5}{3}$$
Note that we do not need to choose the second factor since the order of the two prime factors does not matter.

*Two primes are in one factor, two of the remaining primes are in another factor, and the remaining factor is prime.
$$\frac{1}{2}\binom{5}{2}\binom{3}{2}\binom{1}{1}$$
where we multiply by $1/2$ since choosing $2$ and $3$ for the first factor and $5$ and $7$ for the second factor yields the same factorization as choosing $5$ and $7$ for the first factor and $2$ and $3$ for the second factor.  Alternatively, we choose the prime factor, then choose which of the three primes will be paired with the smallest remaining prime.
$$\binom{5}{1}\binom{3}{1}$$


Observe that the cases are disjoint, so the number of ways of factoring $2310$ into three factors is found by summing the above results.
