Math Problem Statement

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

Solution

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

The expression is:

5C24C112C3\frac{{^5C_2 \cdot ^4C_1}}{{^{12}C_3}}

Step 1: Evaluate Each Combination Term

  1. Calculate 5C2^5C_2: 5C2=5!2!(52)!=5421=10^5C_2 = \frac{5!}{2!(5-2)!} = \frac{5 \cdot 4}{2 \cdot 1} = 10

  2. Calculate 4C1^4C_1: 4C1=4!1!(41)!=41=4^4C_1 = \frac{4!}{1!(4-1)!} = \frac{4}{1} = 4

  3. Calculate 12C3^{12}C_3: 12C3=12!3!(123)!=121110321=13206=220^{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:

5C24C112C3=104220=40220\frac{{^5C_2 \cdot ^4C_1}}{{^{12}C_3}} = \frac{10 \cdot 4}{220} = \frac{40}{220}

Step 3: Simplify the Fraction

40220=211\frac{40}{220} = \frac{2}{11}

Final Answer:

5C24C112C3=211\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 12C4^{12}C_4 instead of 12C3^{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.

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Math Problem Analysis

Mathematical Concepts

Combinatorics
Combinations
Fraction Simplification

Formulas

Combination formula: \( ^nC_r = \frac{n!}{r!(n - r)!} \)

Theorems

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Suitable Grade Level

Grades 10-12