Let $$A = \{a, b, c\}$$ and $$B = \{1, 2, 3, 4\}$$. Then the number of elements in the set $$C = \{f : A \to B \mid 2 \in f(A)$$ and $$f$$ is not one-one$$\}$$ is...

We have $$A = \{a,\; b,\; c\}$$ so $$|A| = 3$$, and $$B = \{1,\; 2,\; 3,\; 4\}$$ so $$|B| = 4$$. A function $$f : A \to B$$ assigns to every element of $$A$$ exactly one element of $$B$$.

First, recall that the total number of all possible functions from a set with $$m$$ elements to a set with $$n$$ elements is $$n^{\,m}$$ because each element of the domain has $$n$$ independent choices. Here, $$m = 3$$ and $$n = 4$$, so the total number of functions is

$$4^{\,3} = 64.$$

We are interested only in those functions which satisfy two simultaneous conditions:

(i) $$2 \in f(A)\;,$$ meaning that at least one element of $$A$$ is mapped to $$2$$.

(ii) $$f$$ is not one-one (not injective), i.e. at least two elements of $$A$$ share the same image.

We shall count functions satisfying (i) first, then remove those that violate (ii). Let us denote by $$N_1$$ the number of functions satisfying (i) and by $$N_{1,\,\text{inj}}$$ the number of those functions which are also injective. The desired count will therefore be

$$N = N_1 \;-\; N_{1,\,\text{inj}}.$$

Counting $$N_1$$ - functions with $$2$$ in the image.

Consider the complementary set of functions where $$2$$ never appears. For every element of $$A$$ we then have only $$\{1,3,4\}$$ available, that is $$3$$ choices per element. Hence the number of such functions is

$$3^{\,3} = 27.$$

By subtraction, the number of functions that do include $$2$$ is

$$N_1 = 64 \;-\; 27 = 37.$$

Counting $$N_{1,\,\text{inj}}$$ - injective functions with $$2$$ in the image.

A function from a set of $$3$$ elements to a set of $$4$$ elements is injective exactly when it assigns three distinct images. The total number of injective functions is therefore the number of permutations of any $$3$$ distinct elements chosen from $$4$$, which is the permutation number

$$P(4,3) = 4 \times 3 \times 2 = 24.$$

Among these, let us identify how many avoid the value $$2$$ entirely; such functions must use the three values $$1,3,4$$. There is only one way to choose these three values, and they can be arranged in $$P(3,3) = 3 \times 2 \times 1 = 6$$ ways. Hence

$$\text{Injective functions without }2 = 6.$$

Consequently, the injective functions that do contain $$2$$ are

$$N_{1,\,\text{inj}} = 24 \;-\; 6 = 18.$$

Final subtraction.

We now exclude these $$18$$ injective functions from the $$37$$ functions that contain $$2$$:

$$N = 37 \;-\; 18 = 19.$$

So, the answer is $$19$$.