Number of Derangements

1. Partition the Domain Sets
The set $$S$$ contains the first $$9$$ natural numbers:
$$S = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$$

Event $$E_2$$ specifies that the set of prime numbers maps onto itself:
$$P = \{2, 3, 5, 7\} \implies f(P) = P$$

Since the function $$f$$ is a bijection, mapping the $$4$$ prime numbers to themselves implies that the remaining $$5$$ non-prime elements must also map onto themselves:
$$NP = \{1, 4, 6, 8, 9\} \implies f(NP) = NP$$

2. Formulate Derangement Constraints
Event $$E_1$$ states that the function forms a complete derangement of the set $$S$$, which means no element maps to itself:
$$f(i) \neq i \quad \forall i \in S$$

To satisfy both $$E_1$$ and $$E_2$$ simultaneously:
1. The prime elements in $$P$$ must be deranged exclusively within the set $$P$$.
2. The non-prime elements in $$NP$$ must be deranged exclusively within the set $$NP$$.

The number of ways to completely derange a set of size $$n$$ is given by the subfactorial formula $$!n$$:
$$!n = n! \sum_{k=0}^{n} \frac{(-1)^k}{k!}$$

3. Calculate Independent Components
Compute the derangements for both partitioned subsets independently:

For Prime Elements ($$\vert{}P\vert{} = 4$$):
$$!4 = 4! \left( 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} \right) = 12 - 4 + 1 = 9$$

For Non-Prime Elements ($$\vert{}NP\vert{} = 5$$):
$$!5 = 5! \left( 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120} \right) = 60 - 20 + 5 - 1 = 44$$

4. Find the Total Product
By the fundamental counting principle, the total number of valid bijective functions $$N$$ is the product of the number of choices for each independent partition:
$$N = !4 \times !5$$
$$N = 9 \times 44 = 396$$

The total number of unique bijective functions $$f(x)$$ satisfying both conditions is $$396$$