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To solve this expression, we’ll break it down by evaluating each combination (denoted as \( C \)) term and then performing the necessary multiplication and division.

The expression is:

\[
\frac{{^5C_2 \cdot ^4C_1}}{{^{12}C_3}}
\]

### Step 1: Evaluate Each Combination Term
1. **Calculate \( ^5C_2 \):**
   \[
   ^5C_2 = \frac{5!}{2!(5-2)!} = \frac{5 \cdot 4}{2 \cdot 1} = 10
   \]

2. **Calculate \( ^4C_1 \):**
   \[
   ^4C_1 = \frac{4!}{1!(4-1)!} = \frac{4}{1} = 4
   \]

3. **Calculate \( ^{12}C_3 \):**
   \[
   ^{12}C_3 = \frac{12!}{3!(12-3)!} = \frac{12 \cdot 11 \cdot 10}{3 \cdot 2 \cdot 1} = \frac{1320}{6} = 220
   \]

### Step 2: Substitute and Simplify
Now, substitute these values into the original expression:

\[
\frac{{^5C_2 \cdot ^4C_1}}{{^{12}C_3}} = \frac{10 \cdot 4}{220} = \frac{40}{220}
\]

### Step 3: Simplify the Fraction
\[
\frac{40}{220} = \frac{2}{11}
\]

### Final Answer:
\[
\frac{{^5C_2 \cdot ^4C_1}}{{^{12}C_3}} = \frac{2}{11}
\]

Would you like further explanation or have any questions?

---

1. What are combinations and how do they differ from permutations?
2. How is the formula for combinations derived?
3. What are other real-world applications of combinations?
4. How would the answer change if we used \( ^{12}C_4 \) instead of \( ^{12}C_3 \)?
5. Can the combination formula be used for non-integer values?

**Tip:** For simplification, always check if the numerator and denominator share any common factors. This can save time when reducing fractions.

Simplify \( \frac{{^5C_2 \cdot ^4C_1}}{{^{12}C_3}} \) and express the answer as an integer or a simplified fraction.

Explore how to simplify the combinatorial expression \( \frac{{^5C_2 \cdot ^4C_1}}{{^{12}C_3}} \) with step-by-step calculations. This problem demonstrates the use of combination formulas and fraction simplification, suitable for high school students in Grades 10-12.

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