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College Algebra
7th Edition
ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter9: Counting And Probability
Section9.1: Counting
Problem 1E: The Fundamental Counting Principle says that if one event can occur in m ways and a second event can...
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Contingency Table
A contingency table can be defined as the visual representation of the relationship between two or more categorical variables that can be evaluated and registered. It is a categorical version of the scatterplot, which is used to investigate the linear relationship between two variables. A contingency table is indeed a type of frequency distribution table that displays two variables at the same time.
Binomial Distribution
Binomial is an algebraic expression of the sum or the difference of two terms. Before knowing about binomial distribution, we must know about the binomial theorem.
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![**Question 10**
Compute \( \binom{5}{2} \)
**Explanation:**
The expression \( \binom{5}{2} \) represents a binomial coefficient, also known as "5 choose 2," which is used in combinatorics to determine the number of ways to choose 2 elements from a set of 5 elements without regard to the order of selection.
The formula to compute \( \binom{n}{k} \) is:
\[
\binom{n}{k} = \frac{n!}{k!(n-k)!}
\]
In this case:
- \( n = 5 \)
- \( k = 2 \)
Therefore:
\[
\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4 \times 3!}{2 \times 1 \times 3!} = \frac{20}{2} = 10
\]
So, there are 10 ways to choose 2 elements from a set of 5 elements.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa8a3b459-3794-4f09-9dad-faeab1970da3%2Fa2b467de-ceda-429b-8f47-31cd4f739a20%2Fflpxec3_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 10**
Compute \( \binom{5}{2} \)
**Explanation:**
The expression \( \binom{5}{2} \) represents a binomial coefficient, also known as "5 choose 2," which is used in combinatorics to determine the number of ways to choose 2 elements from a set of 5 elements without regard to the order of selection.
The formula to compute \( \binom{n}{k} \) is:
\[
\binom{n}{k} = \frac{n!}{k!(n-k)!}
\]
In this case:
- \( n = 5 \)
- \( k = 2 \)
Therefore:
\[
\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4 \times 3!}{2 \times 1 \times 3!} = \frac{20}{2} = 10
\]
So, there are 10 ways to choose 2 elements from a set of 5 elements.
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Step 1
We have in general the formula for combination as,
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