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The problem asks how many functions \( g : \{a, b, c\} \to \{1, 2, 3, 4, 5\} \) are such that the composition \( f \circ g \) is a bijection on \( \{a, b, c\} \).

### Understanding the problem:
- We are given a function \( f : \{1, 2, 3, 4, 5\} \to \{a, b, c\} \) where \( f(1) = a \), \( f(2) = a \), \( f(3) = b \), \( f(4) = c \), and \( f(5) = c \).
- We are also defining a function \( g : \{a, b, c\} \to \{1, 2, 3, 4, 5\} \), and we want the composition \( f \circ g \) to be a bijection from \( \{a, b, c\} \) to itself.

### Key Points:
- \( f \circ g \) means that for each element \( x \in \{a, b, c\} \), we first apply \( g \) to \( x \), then apply \( f \) to the result of \( g(x) \). So, for each element \( x \in \{a, b, c\} \), \( f(g(x)) \) should return \( a, b, \) or \( c \).
- Since \( f \circ g \) must be a bijection on \( \{a, b, c\} \), each element of \( \{a, b, c\} \) must be mapped to a unique element of \( \{a, b, c\} \) by the composition.

### Step-by-Step Solution:

1. **Behavior of \( f \):**
   - \( f(1) = a \) and \( f(2) = a \): Both 1 and 2 are mapped to \( a \).
   - \( f(3) = b \): 3 is mapped to \( b \).
   - \( f(4) = c \) and \( f(5) = c \): Both 4 and 5 are mapped to \( c \).

2. **Conditions for \( g \):**
   - To make \( f \circ g \) a bijection, each \( x \in \{a, b, c\} \) must be mapped to one of the preimages of \( x \) under \( f \). This means:
     - \( g(a) \) must be either 1 or 2 (since \( f(1) = a \) and \( f(2) = a \)).
     - \( g(b) \) must be 3 (since \( f(3) = b \)).
     - \( g(c) \) must be either 4 or 5 (since \( f(4) = c \) and \( f(5) = c \)).

3. **Counting the possibilities:**
   - For \( g(a) \), we have 2 choices: either 1 or 2.
   - For \( g(b) \), there is only 1 choice: 3.
   - For \( g(c) \), we have 2 choices: either 4 or 5.

4. **Total number of functions:**
   - The total number of possible functions \( g \) is the product of these choices:
     \[
     2 \times 1 \times 2 = 4
     \]

Thus, there are 4 possible functions \( g \) such that \( f \circ g \) is a bijection.

The correct answer is **4**.

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Would you like a more detailed explanation or have any further questions? Here are five related questions that might help expand this understanding:

1. How would the problem change if \( f \) had different mappings, for example, if \( f(1) = b \)?
2. What are the general conditions for a function composition to be bijective?
3. How many bijections exist from a set of 4 elements to itself?
4. Can you always create a bijection if the sets involved have different cardinalities?
5. How does the concept of injectivity differ from bijectivity?

**Tip:** When dealing with bijective functions, always ensure that every element in both the domain and codomain has a unique match. This can simplify understanding the restrictions on function composition.

Let f: {1, 2, 3, 4, 5} → {a, b, c} be such that f(1) = a, f(2) = a, f(3) = b, f(4) = c, f(5) = c. How many functions g: {a, b, c} → {1, 2, 3, 4, 5} are there such that the function f ∘ g: {a, b, c} → {a, b, c} is a bijective function on the set {a, b, c}?

Explore how to determine the number of functions g: {a, b, c} → {1, 2, 3, 4, 5} that result in a bijective composite function f ∘ g: {a, b, c} → {a, b, c}. This problem involves function composition and bijective mapping principles, suitable for students in Grades 10-12.

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