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Evaluate \( { }_{11} \mathrm{C}_{3} \). \( { }_{11} \mathrm{C}_{3}=\square \) (Simplify your answer. Type an integer or a fraction

Ask by Rowe Ryan. in the United States
Mar 11,2025

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Tutor-Verified Answer

Answer

\[ { }_{11}C_{3} = 165 \]

Solution

We start by writing the combination formula: \[ { }_{n}C_{k}=\frac{n!}{k!(n-k)!} \] Substitute \( n=11 \) and \( k=3 \): \[ { }_{11}C_{3}=\frac{11!}{3!(11-3)!}=\frac{11!}{3!8!} \] Next, simplify by writing the factorial \( 11! \) as: \[ 11!=11\times10\times9\times8! \] This allows us to cancel the \( 8! \) in the numerator and denominator: \[ { }_{11}C_{3}=\frac{11\times10\times9\times8!}{3!\times8!}=\frac{11\times10\times9}{3!} \] Recall that: \[ 3! = 3\times2\times1=6 \] Now, compute the multiplication in the numerator: \[ 11\times10\times9=990 \] Finally, divide by \( 6 \): \[ { }_{11}C_{3}=\frac{990}{6}=165 \] Thus, the answer is: \[ { }_{11}C_{3}=165 \]

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To evaluate \( { }_{11} \mathrm{C}_{3} \), we use the formula for combinations: \[ { }_{n} \mathrm{C}_{r} = \frac{n!}{r!(n-r)!} \] Substituting \( n = 11 \) and \( r = 3 \): \[ { }_{11} \mathrm{C}_{3} = \frac{11!}{3!(11-3)!} = \frac{11!}{3! \cdot 8!} \] Next, we can simplify \( 11! \): \[ 11! = 11 \times 10 \times 9 \times 8! \] So we have: \[ { }_{11} \mathrm{C}_{3} = \frac{11 \times 10 \times 9 \times 8!}{3! \times 8!} = \frac{11 \times 10 \times 9}{3!} \] Calculating \( 3! = 3 \times 2 \times 1 = 6 \): Now plug this into our equation: \[ { }_{11} \mathrm{C}_{3} = \frac{11 \times 10 \times 9}{6} \] Calculating the numerator: \[ 11 \times 10 = 110 \] \[ 110 \times 9 = 990 \] Now divide by 6: \[ \frac{990}{6} = 165 \] Therefore, the final answer is: \[ { }_{11} \mathrm{C}_{3} = 165 \]

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