The Number Of Ways In Which 10 Candidates A 1 , A 2 , . . . , A 10 A_1,A_2,...,A_{10} A 1 ​ , A 2 ​ , ... , A 10 ​ Can Be Ranked So That A 1 A_1 A 1 ​ Is Always Above A 2 A_2 A 2 ​ , Is:

Introduction

In this article, we will explore the problem of ranking 10 candidates, denoted as $A_1, A_2, ..., A_{10}$, such that $A_1$ is always above $A_2$. This problem falls under the category of combinatorics, permutations, and combinations. We will use a step-by-step approach to find the number of ways to rank the candidates under the given condition.

Understanding the Problem

The problem requires us to find the number of ways to rank 10 candidates, with the condition that $A_1$ is always above $A_2$. This means that $A_1$ must be ranked higher than $A_2$ in all possible rankings.

Approach

To solve this problem, we can consider $A_1$ and $A_2$ as a single object, denoted as $A_1A_2$. This is because the relative position of $A_1$ and $A_2$ is fixed, with $A_1$ always above $A_2$. By considering them as a single object, we can reduce the problem to finding the number of ways to rank 9 objects, which is a more manageable task.

Step 1: Counting the Number of Ways to Rank 9 Objects

We can use the concept of permutations to count the number of ways to rank 9 objects. The number of permutations of $n$ objects is given by $n!$, where $n! = n \times (n-1) \times (n-2) \times ... \times 2 \times 1$. In this case, we have 9 objects, so the number of permutations is $9!$.

Step 2: Accounting for the Fixed Position of A1 and A2

Since $A_1$ and $A_2$ are considered as a single object, we need to account for the fixed position of $A_1$ and $A_2$. This means that for every permutation of the 9 objects, we need to consider the position of $A_1$ and $A_2$ as fixed. This reduces the number of permutations by a factor of 2, since $A_1$ and $A_2$ can be swapped.

Step 3: Calculating the Final Answer

Taking into account the fixed position of $A_1$ and $A_2$, the number of ways to rank the 10 candidates is given by:

$$\frac{9!}{2}$$

This is because we divide the total number of permutations by 2 to account for the fixed position of $A_1$ and $A_2$.

Conclusion

In this article, we have explored the problem of ranking 10 candidates, with the condition that $A_1$ is always above $A_2$. We used a step-by-step approach to find the number of ways to rank the candidates under the given condition. By considering $A_1$ and $A_2$ as a single object, we reduced the problem to finding the number of ways to rank 9 objects. We then accounted for the fixed position of $A_1$ and $A_2$ and calculated the final answer.

Final Answer

The number ways to rank 10 candidates, with $A_1$ always above $A_2$, is given by:

$$\frac{9!}{2} = \frac{362880}{2} = 181440$$

This is the final answer to the problem.

Related Problems

This problem is related to other combinatorics and permutations problems, such as:

• Finding the number of ways to rank $n$ objects, with the condition that $A_1$ is always above $A_2$.
• Finding the number of ways to arrange $n$ objects, with the condition that $A_1$ is always above $A_2$.

These problems can be solved using similar techniques and approaches.

References

• [1] "Combinatorics: Topics, Techniques, Algorithms" by Peter J. Cameron
• [2] "Permutations and Combinations" by Arthur T. Benjamin

Introduction

In our previous article, we explored the problem of ranking 10 candidates, denoted as $A_1, A_2, ..., A_{10}$, such that $A_1$ is always above $A_2$. We used a step-by-step approach to find the number of ways to rank the candidates under the given condition. In this article, we will answer some frequently asked questions related to this problem.

Q: What is the significance of considering A1 and A2 as a single object?

A: Considering $A_1$ and $A_2$ as a single object is a common technique used in combinatorics to simplify complex problems. By doing so, we reduce the number of objects to be ranked from 10 to 9, making the problem more manageable.

Q: Why do we need to account for the fixed position of A1 and A2?

A: We need to account for the fixed position of $A_1$ and $A_2$ because they are considered as a single object. This means that for every permutation of the 9 objects, we need to consider the position of $A_1$ and $A_2$ as fixed. This reduces the number of permutations by a factor of 2, since $A_1$ and $A_2$ can be swapped.

Q: How do we calculate the number of ways to rank the 10 candidates?

A: To calculate the number of ways to rank the 10 candidates, we use the formula:

$$\frac{9!}{2}$$

This formula takes into account the fixed position of $A_1$ and $A_2$ and gives us the total number of ways to rank the 10 candidates.

Q: What is the final answer to the problem?

A: The final answer to the problem is:

$$\frac{9!}{2} = \frac{362880}{2} = 181440$$

This is the total number of ways to rank the 10 candidates, with $A_1$ always above $A_2$.

Q: Are there any related problems that can be solved using similar techniques?

A: Yes, there are several related problems that can be solved using similar techniques. Some examples include:

• Finding the number of ways to rank $n$ objects, with the condition that $A_1$ is always above $A_2$.
• Finding the number of ways to arrange $n$ objects, with the condition that $A_1$ is always above $A_2$.

These problems can be solved using similar techniques and approaches.

Q: What are some common applications of combinatorics and permutations?

A: Combinatorics and permutations have numerous applications in various fields, including:

• Computer science: Combinatorics and permutations are used in algorithms and data structures to solve complex problems.
• Statistics: Combinatorics and permutations are used in statistical analysis to model and analyze data.
• Biology: Combinatorics and permutations are used in bioinformatics to analyze and model biological systems.
• Economics: Combinator and permutations are used in economic modeling to analyze and predict market trends.

These are just a few examples of the many applications of combinatorics and permutations.

Conclusion

In this article, we have answered some frequently asked questions related to the problem of ranking 10 candidates, with $A_1$ always above $A_2$. We have used a step-by-step approach to explain the solution and provide examples of related problems. We hope that this article has been helpful in understanding the concepts of combinatorics and permutations.