Factorization Possibilities

This misses out two scenarios, namely (i) two pairs of prime factors, e.g., $(2*3)*(5*7)*11 = 6*35*11$ , ( $15$ cases), and (ii) one triple and one pair of prime factors, e.g., $(2*3*5)*(7*11) = 30*77$ , ( $10$ cases). It also over-counts with regards to the scenario where no prime factors are combined; there is only one case to count rather that $\binom{5}{1} = 5$ cases. Thus we end up with $15 + 10 - 4 = 21$ more cases, for a total of $31 + 21 = 52$ cases, as Arjen has stated.

You could click on the link I provided :)

In short, the Bell number $B_n$ is the number of partitions of a set with $n$ elements. In this case, we count the partitions of the set $\{2, 3, 5, 7, 11\}$ , so the answer is $B_5$ .

A basic recursion formula for the calculation of $B_n$ is $B_0 = 1;\ \ B_{n+1} = \sum_{k = 0}^n \left(\begin{array}{c} n \\ k\end{array}\right)\:B_k.$ The first few Bell numbers are $B_0 = 1;\ \ B_1 = 1;\ \ B_2 = 2;\ \ B_3 = 5;\ \ B_4 = 15;\ \ B_5 = 52;\ \ B_6 = 203.$

distinct objects into identical bins

Try this out : -link-

For $n = 2$ the answer is $\left\lceil \frac {\delta(N)} 2 \right\rceil,$ where $\delta(N)$ is the number of divisors of the given number $N$ . Moreover, if $N = \prod p_i^{e_i},$ where the $p_i$ are the distinct prime factors of $N$ , then $\delta(N) = \prod (e_i+1).$

For your example, $\delta(4620) = 3\cdot 2\cdot 2\cdot 2\cdot 2 = 48,$ which means that there are 24 ways to factorize 4620 as the product of two factors.

Rationale: if $d$ is a divisor of $N$ , then $N = d\cdot \frac N d$ is a factorization; and all factorizations of $N$ are of this form. However, we now have all factorizations double, e.g. $4620 = 60\cdot 77$ and $4620 = 77\cdot 60$ ; therefore we divide by two. The only exception is when $\delta(N)$ is odd, which means that $N$ is a perfect square. In that case the factorization $N = \sqrt N \cdot \sqrt N$ should be excluded from the division by two; this I accomplish with the ceiling function $\lceil\cdot\rceil$ .

@Vladimir Smith See this , you will see that it is a known result that this works for square free numbers only.