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
 A: 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 $$25+15+1$$
A: You have two ways of "splitting" these factors in 3 groups : 1-2-2 or 1-1-3.
In the first case, you might solve the problem by looking at the number of ways of ordering your 5 factors (and so choosing your first factor as the first number in your order, the second as the product of the two next numbers, and so on), and then dividing by 2 for the number of ways of ordering your first pair, dividing by 2 for the number of ways of ordering your second pair, and finally dividing by 2 as the 2 pairs can be switched. This gives you $\frac{5!}{2^3}=15$.
You might now use a similar logic in the 2nd case, and you will obtain $\frac{5!}{2\cdot 3!}=10$, as you divide by 2 to remove the repetitions of the order of the 2 individual factors, and $3!$ to remove the ordering inside the group of 3.
And so, you get $25$ possible ways for this case.
A: 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.
A: The three cases:
$1)$ Two factors are ones: total options: $\fbox{1}$.
$2)$ One factor is one: two partitions: ${\{1,4\}} \Rightarrow C^5_1=5$ options and ${\{2,3\}} \Rightarrow C^5_2=10$ options, hence total options: $\fbox{15}$  
$3)$ No factor is one: two partitions: ${\{1,2,2\}} \Rightarrow C^5_1\cdot \frac{C^4_2}{2}=15$ options and ${\{1,1,3\}} \Rightarrow C^5_1\cdot \frac{C^4_1}{2}=10$ options, hence total options: $\fbox{25}.$ 
Grand total: $\fbox{41}$ options.
