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
Best Answer
The number of ways to place $n$ different objects into $k$ indistinguishable bags is $$\left\{{n\atop k}\right\}$$which is a Stirling number of the second kind.
In this case, the objects are the five prime factors of $2310$, and the bags are our three possibly composite factors greater than $1$: $$\left\{{5\atop 3}\right\}+\left\{{5\atop 2}\right\}+\left\{{5\atop 1}\right\}$$
Your calculation of the last two terms was correct. To find $\left\{{5\atop 3}\right\}$, we can use inclusion-exclusion.
Represent the placement of objects into labelled bags by a string of length $5$, e.g. $21312$ means that the first object goes into bag $2$, the second into bag $1$, the third into bag $3$, etc.
There are $3^5$ such strings; of those, $\binom31\cdot2^5$ miss out at least one bag, while $\binom32\cdot1^5$ miss out two.
Thus there are $3^5-3\cdot2^5+3\cdot1^5=150$ strings with all the digits from $1$ to $3$.
However, the bags aren't supposed to have labels, so we divide by the number of orders in which $1,2,3$ can appear: i.e. $3!$ to get the answer $\left\{{5\atop 3}\right\}=25$. Thus the final answer is $$25+15+1$$