Evaluate the expression. C(11,4)
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...
Related questions
Question
![---
### 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.
---](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F35ad263b-c31f-4196-aa81-a0a5a787610e%2Fd3759bb9-c804-46b3-8d69-03ba6339abfd%2Fd55x349_processed.jpeg&w=3840&q=75)
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.
---
Expert Solution

This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps

Recommended textbooks for you

A First Course in Probability (10th Edition)
Probability
ISBN:
9780134753119
Author:
Sheldon Ross
Publisher:
PEARSON


A First Course in Probability (10th Edition)
Probability
ISBN:
9780134753119
Author:
Sheldon Ross
Publisher:
PEARSON
