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Use a calculator to evaluate \( { }_{27} \mathrm{C}_{12} \). \( { }_{27} \mathrm{C}_{12}=\square \) (Simplify your answer. Type an integer or a fraction.)

Ask by Long Moran. in the United States
Mar 10,2025

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Tutor-Verified Answer

Answer

\[ {27 \choose 12} = 17383860 \]

Solution

We start with the definition of the binomial coefficient, \[ {27 \choose 12} = \frac{27!}{12! \, 15!}. \] This can be written as a product of 12 factors: \[ {27 \choose 12} = \frac{27 \times 26 \times 25 \times \cdots \times 16}{12 \times 11 \times 10 \times \cdots \times 1}. \] One way to simplify this is to write \[ {27 \choose 12} = \frac{16}{1} \times \frac{17}{2} \times \frac{18}{3} \times \frac{19}{4} \times \frac{20}{5} \times \frac{21}{6} \times \frac{22}{7} \times \frac{23}{8} \times \frac{24}{9} \times \frac{25}{10} \times \frac{26}{11} \times \frac{27}{12}. \] Let’s simplify step by step. 1. Notice that \[ \frac{18}{3} = 6, \quad \frac{20}{5} = 4, \quad \frac{21}{6} = \frac{7}{2}, \quad \frac{24}{9} = \frac{8}{3}, \quad \frac{25}{10} = \frac{5}{2}, \quad \frac{27}{12} = \frac{9}{4}. \] 2. Replace these in the product: \[ 16 \times \frac{17}{2} \times 6 \times \frac{19}{4} \times 4 \times \frac{7}{2} \times \frac{22}{7} \times \frac{23}{8} \times \frac{8}{3} \times \frac{5}{2} \times \frac{26}{11} \times \frac{9}{4}. \] 3. Now simplify by canceling common factors: - \( \frac{7}{2} \times \frac{22}{7} = \frac{22}{2} = 11 \). - \( \frac{23}{8} \times \frac{8}{3} = \frac{23}{3} \). - Also, note that the factor \( \frac{26}{11} \) multiplied by the \(11\) obtained becomes \(26\). 4. Group the remaining factors: - The integer factors: \(16 \times 6 \times 4\). - The remaining fractions: \(\frac{17}{2} \times \frac{19}{4} \times \frac{23}{3} \times \frac{5}{2} \times \frac{9}{4}\) and the factor \(26\). First, combine the integers: \[ 16 \times 6 \times 4 = 384. \] 5. The product now is: \[ 384 \times \frac{17}{2} \times \frac{19}{4} \times \frac{23}{3} \times \frac{5}{2} \times \frac{9}{4} \times 26. \] 6. Combine the fractions. Compute the combined denominator and numerator: - The combined denominator is: \[ 2 \times 4 \times 3 \times 2 \times 4 = 192. \] - The combined numerator from the fractions is: \[ 17 \times 19 \times 23 \times 5 \times 9. \] Now, include the factor \(26\) with these factors. 7. So we have: \[ {27 \choose 12} = 384 \times \frac{17 \times 19 \times 23 \times 5 \times 26 \times 9}{192}. \] 8. Notice that \( \frac{384}{192} = 2 \). Therefore, \[ {27 \choose 12} = 2 \times (17 \times 19 \times 23 \times 5 \times 26 \times 9). \] 9. Compute the product step by step: - \(17 \times 19 = 323\), - \(323 \times 23 = 7429\), - \(7429 \times 5 = 37145\), - \(37145 \times 26 = 37145 \times 20 + 37145 \times 6 = 742900 + 222870 = 965770\), - \(965770 \times 9 = 8\,691\,930\). 10. Finally, multiply by 2: \[ 2 \times 8\,691\,930 = 17\,383\,860. \] Thus, the evaluated answer is: \[ {27 \choose 12} = 17383860. \]

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To evaluate \( { }_{27} \mathrm{C}_{12} \), we use the combination formula: \[ { }_{n} \mathrm{C}_{r} = \frac{n!}{r!(n-r)!} \] For our problem, \( n = 27 \) and \( r = 12 \): \[ { }_{27} \mathrm{C}_{12} = \frac{27!}{12!(27-12)!} = \frac{27!}{12! \cdot 15!} \] Using a calculator, we find: \[ { }_{27} \mathrm{C}_{12} = \frac{27 \times 26 \times 25 \times 24 \times 23 \times 22 \times 21 \times 20 \times 19 \times 18 \times 17 \times 16}{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} \] Calculating that gives us: \[ { }_{27} \mathrm{C}_{12} = 17383860 \] Therefore, \( { }_{27} \mathrm{C}_{12} = 17383860 \).

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