Introduction:
The symmetric group S3 consists of all possible permutations of three elements. A subgroup is said to be normal if it is invariant under conjugation by elements of the group. In this case, we need to determine the number of distinct normal subgroups of S3.

Explanation:
To find the number of distinct normal subgroups of S3, we can consider the order of the subgroups. The possible orders of subgroups of S3 are 1, 2, 3, and 6.

Order 1:
A subgroup of order 1 is simply the identity element {e}. Since the identity element is in the center of any group, it is automatically a normal subgroup.

Order 2:
There are three subgroups of order 2 in S3, namely {(1 2)}, {(1 3)}, and {(2 3)}. To check if they are normal, we need to determine if they are invariant under conjugation. Conjugation of a subgroup H by an element g is defined as gHg^-1.

Let's take the subgroup {(1 2)} as an example. To check if it is normal, we need to check if for any g in S3, g{(1 2)}g^-1 is contained in {(1 2)}.

Let's consider g = (1 2 3). We have (1 2 3){(1 2)}(1 2 3)^-1 = (1 2 3)(1 2)(3 2 1) = (1 3), which is not in {(1 2)}. Therefore, {(1 2)} is not a normal subgroup.

Similarly, we can check that {(1 3)} and {(2 3)} are not normal subgroups.

Order 3:
There is only one subgroup of order 3 in S3, namely {(1 2 3)}. Since there are only three elements in S3, every element is of order 2 or 3. Therefore, {(1 2 3)} is the only subgroup of order 3 and it is normal.

Order 6:
The only subgroup of order 6 in S3 is the entire group itself, S3. Since every group is normal in itself, S3 is a normal subgroup.

Conclusion:
In summary, the number of distinct normal subgroups of S3 is 3. They are the trivial subgroup of order 1, the subgroup {(1 2 3)} of order 3, and the entire group S3 of order 6.