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Evaluate the expression. C(11,4)

Evaluate the expression. C(11,4)

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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--- ### Evaluating Combinations **Task:** Evaluate the expression. \[ C(11,4) \] \[ C(11,4) = \boxed{\quad} \] --- ### Explanation This problem involves calculating a combination, denoted as \( C(n, k) \), which represents the number of ways to choose \( k \) elements from a set of \( n \) elements without regard to the order of selection. The formula for combinations is: \[ C(n, k) = \frac{n!}{k!(n-k)!} \] For this expression, \( n = 11 \) and \( k = 4 \). 1. **Calculate Factorials:** - \( 11! = 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \) - \( 4! = 4 \times 3 \times 2 \times 1 \) - \( (11-4)! = 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \) 2. **Substitute and Simplify:** - \( C(11, 4) = \frac{11!}{4! \times 7!} \) 3. **Result:** - Evaluate the expression to find the number of combinations. This calculation is often used in probability, statistics, and many fields of mathematics. ---
Transcribed Image Text:--- ### Evaluating Combinations **Task:** Evaluate the expression. \[ C(11,4) \] \[ C(11,4) = \boxed{\quad} \] --- ### Explanation This problem involves calculating a combination, denoted as \( C(n, k) \), which represents the number of ways to choose \( k \) elements from a set of \( n \) elements without regard to the order of selection. The formula for combinations is: \[ C(n, k) = \frac{n!}{k!(n-k)!} \] For this expression, \( n = 11 \) and \( k = 4 \). 1. **Calculate Factorials:** - \( 11! = 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \) - \( 4! = 4 \times 3 \times 2 \times 1 \) - \( (11-4)! = 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \) 2. **Substitute and Simplify:** - \( C(11, 4) = \frac{11!}{4! \times 7!} \) 3. **Result:** - Evaluate the expression to find the number of combinations. This calculation is often used in probability, statistics, and many fields of mathematics. ---
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