The largest $$n \in \mathbb{N}$$ such that $$3^n$$ divides 50! is:

To find the highest power of a prime $$p$$ that divides $$m!$$, we use Legendre’s formula:
$$\nu_p(m!) = \left\lfloor\frac{m}{p}\right\rfloor + \left\lfloor\frac{m}{p^2}\right\rfloor + \left\lfloor\frac{m}{p^3}\right\rfloor + \cdots$$
where $$\nu_p(m!)$$ denotes the exponent of $$p$$ in the prime-factorisation of $$m!$$.

Here, $$p = 3$$ and $$m = 50$$. We compute each term until the quotient becomes zero:

First term: $$\left\lfloor\frac{50}{3}\right\rfloor = 16$$

Second term: $$\left\lfloor\frac{50}{9}\right\rfloor = 5$$

Third term: $$\left\lfloor\frac{50}{27}\right\rfloor = 1$$

Fourth term: $$\left\lfloor\frac{50}{81}\right\rfloor = 0$$ (and all further terms are also $$0$$)

Add the non-zero contributions:
$$\nu_3(50!) = 16 + 5 + 1 = 22$$

Therefore the greatest integer $$n$$ such that $$3^n$$ divides $$50!$$ is $$n = 22$$.

Hence, the correct option is Option B.