2) A jar consists of 7 red marbles and 11 green marbles a) How man selections of 8 marbles are there b) How many selections of 8 marbles are there where 3 are red and 5 are green
2) A jar consists of 7 red marbles and 11 green marbles a) How man selections of 8 marbles are there b) How many selections of 8 marbles are there where 3 are red and 5 are green
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Permutations and Combinations
If there are 5 dishes, they can be relished in any order at a time. In permutation, it should be in a particular order. In combination, the order does not matter. Take 3 letters a, b, and c. The possible ways of pairing any two letters are ab, bc, ac, ba, cb and ca. It is in a particular order. So, this can be called the permutation of a, b, and c. But if the order does not matter then ab is the same as ba. Similarly, bc is the same as cb and ac is the same as ca. Here the list has ab, bc, and ac alone. This can be called the combination of a, b, and c.
Counting Theory
The fundamental counting principle is a rule that is used to count the total number of possible outcomes in a given situation.
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![### Problem Statement
A jar consists of 7 red marbles and 11 green marbles.
#### Questions:
a) How many selections of 8 marbles are there?
b) How many selections of 8 marbles are there where 3 are red and 5 are green?
### Solutions:
#### Part (a): How many selections of 8 marbles are there?
To determine the number of ways to select 8 marbles from a total of \(7 + 11 = 18\) marbles, we use the combination formula:
\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
where \( n \) is the total number of marbles and \( k \) is the number of marbles to select.
Here, \( n = 18 \) and \( k = 8 \).
Therefore,
\[ \binom{18}{8} = \frac{18!}{8!(18-8)!} \]
#### Part (b): How many selections of 8 marbles are there where 3 are red and 5 are green?
To determine the number of ways to select 3 red marbles from 7 and 5 green marbles from 11, we consider the two selections separately:
1. The number of ways to choose 3 red marbles from 7 is:
\[ \binom{7}{3} = \frac{7!}{3!(7-3)!} \]
2. The number of ways to choose 5 green marbles from 11 is:
\[ \binom{11}{5} = \frac{11!}{5!(11-5)!} \]
The total number of ways to make the combined selection is the product of the above two combinations:
\[ \binom{7}{3} \times \binom{11}{5} \]
### Explanation of Graphs or Diagrams:
There are no graphs or diagrams accompanying this problem statement. The calculations involve combinatorial mathematics, specifically the use of combinations to determine the number of ways to select items. These are solved using combinatorial formulas in a step-by-step manner.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8ae4e595-d9b4-4096-a3fc-f4bee7947b38%2F8ee373f4-ef84-487d-959b-ad22b8eb1bf4%2Fcp61c3o_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
A jar consists of 7 red marbles and 11 green marbles.
#### Questions:
a) How many selections of 8 marbles are there?
b) How many selections of 8 marbles are there where 3 are red and 5 are green?
### Solutions:
#### Part (a): How many selections of 8 marbles are there?
To determine the number of ways to select 8 marbles from a total of \(7 + 11 = 18\) marbles, we use the combination formula:
\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
where \( n \) is the total number of marbles and \( k \) is the number of marbles to select.
Here, \( n = 18 \) and \( k = 8 \).
Therefore,
\[ \binom{18}{8} = \frac{18!}{8!(18-8)!} \]
#### Part (b): How many selections of 8 marbles are there where 3 are red and 5 are green?
To determine the number of ways to select 3 red marbles from 7 and 5 green marbles from 11, we consider the two selections separately:
1. The number of ways to choose 3 red marbles from 7 is:
\[ \binom{7}{3} = \frac{7!}{3!(7-3)!} \]
2. The number of ways to choose 5 green marbles from 11 is:
\[ \binom{11}{5} = \frac{11!}{5!(11-5)!} \]
The total number of ways to make the combined selection is the product of the above two combinations:
\[ \binom{7}{3} \times \binom{11}{5} \]
### Explanation of Graphs or Diagrams:
There are no graphs or diagrams accompanying this problem statement. The calculations involve combinatorial mathematics, specifically the use of combinations to determine the number of ways to select items. These are solved using combinatorial formulas in a step-by-step manner.
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