A surjection is a function that maps every element in the domain to at least one element in the codomain. In other words, every element in the codomain has a pre-image in the domain.

Given:
A = {1, 2, 3, ..., n}
B = {a, b}

Number of Elements in A:
The set A contains n elements.

Number of Elements in B:
The set B contains 2 elements.

Number of Surjections from A to B:
To find the number of surjections from A to B, we need to consider two cases:

1. No element from A maps to 'a' and 'b':
In this case, all the elements in A need to be mapped to 'a' or 'b'. Since each element in A can be mapped to either 'a' or 'b', there are 2 possibilities for each element. Therefore, the total number of surjections in this case is 2^n.

2. At least one element from A maps to 'a' and 'b':
In this case, we subtract the number of surjections where no element maps to 'a' or 'b' from the total number of surjections.

The number of surjections where no element maps to 'a' or 'b' can be calculated using the principle of inclusion-exclusion. There are 2^n total possibilities, out of which:
- nC1 possibilities where 1 element maps to 'a' and 'b'
- nC2 possibilities where 2 elements map to 'a' and 'b'
- nC3 possibilities where 3 elements map to 'a' and 'b'
- ...
- nCn possibilities where all n elements map to 'a' and 'b'

Using the principle of inclusion-exclusion, the number of surjections where at least one element maps to 'a' or 'b' is given by:
Total number of surjections - (nC1 + nC2 + nC3 + ... + nCn)

Therefore, the total number of surjections from A to B is:
2^n - (nC1 + nC2 + nC3 + ... + nCn)

Example:
Let's consider an example where n = 3. In this case, A = {1, 2, 3} and B = {a, b}.

The total number of surjections is 2^3 = 8.
The number of surjections where no element maps to 'a' or 'b' is 0.
Therefore, the number of surjections where at least one element maps to 'a' or 'b' is 8 - (3C1 + 3C2 + 3C3) = 8 - (3 + 3 + 1) = 1.

Hence, the number of surjections from A to B when n = 3 is 1.