Problem 6 How many ways are there for 10 women and six men to stand in a line so that no two men stand next to each other? Hint: First position the women and then consider possible positions for the men.
Problem 6 How many ways are there for 10 women and six men to stand in a line so that no two men stand next to each other? Hint: First position the women and then consider possible positions for the men.
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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![### Problem 6
**Question:**
How many ways are there for 10 women and six men to stand in a line so that no two men stand next to each other?
**Hint:**
First position the women and then consider possible positions for the men.
---
This problem involves combinatorial arrangements and requires the application of permutations while adhering to specific constraints. Below is a step-by-step explanation of how to solve it:
1. **Position the Women:**
First, we arrange the 10 women in a line. The number of ways to arrange 10 women is given by the factorial of 10 (10!).
\[
10! = 3,628,800
\]
2. **Identify Positions for the Men:**
Once the women are positioned, there are 11 possible gaps where the men can be placed (before, between, and after the women). We need to choose 6 out of these 11 gaps to place the men, ensuring that no two men are next to each other. This is a combination problem, expressed as \( \binom{11}{6} \).
\[
\binom{11}{6} = \frac{11!}{6!(11-6)!} = 462
\]
3. **Arrange the Men:**
The number of ways to arrange the 6 men in the chosen gaps is given by the factorial of 6 (6!).
\[
6! = 720
\]
4. **Calculate the Total Number of Ways:**
Multiply the number of ways to arrange the women, choose the gaps for the men, and arrange the men.
\[
10! \times \binom{11}{6} \times 6! = 3,628,800 \times 462 \times 720
\]
Evaluating this product gives:
\[
3,628,800 \times 462 \times 720 = 1,417,236,480,000
\]
So, there are a total of \( 1,417,236,480,000 \) ways for 10 women and 6 men to stand in a line such that no two men are next to each other.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe872339c-719c-4a81-bc7c-37653a99cb41%2F5486522e-3f8e-48de-bd84-97be0135f3eb%2Fvhc09yb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem 6
**Question:**
How many ways are there for 10 women and six men to stand in a line so that no two men stand next to each other?
**Hint:**
First position the women and then consider possible positions for the men.
---
This problem involves combinatorial arrangements and requires the application of permutations while adhering to specific constraints. Below is a step-by-step explanation of how to solve it:
1. **Position the Women:**
First, we arrange the 10 women in a line. The number of ways to arrange 10 women is given by the factorial of 10 (10!).
\[
10! = 3,628,800
\]
2. **Identify Positions for the Men:**
Once the women are positioned, there are 11 possible gaps where the men can be placed (before, between, and after the women). We need to choose 6 out of these 11 gaps to place the men, ensuring that no two men are next to each other. This is a combination problem, expressed as \( \binom{11}{6} \).
\[
\binom{11}{6} = \frac{11!}{6!(11-6)!} = 462
\]
3. **Arrange the Men:**
The number of ways to arrange the 6 men in the chosen gaps is given by the factorial of 6 (6!).
\[
6! = 720
\]
4. **Calculate the Total Number of Ways:**
Multiply the number of ways to arrange the women, choose the gaps for the men, and arrange the men.
\[
10! \times \binom{11}{6} \times 6! = 3,628,800 \times 462 \times 720
\]
Evaluating this product gives:
\[
3,628,800 \times 462 \times 720 = 1,417,236,480,000
\]
So, there are a total of \( 1,417,236,480,000 \) ways for 10 women and 6 men to stand in a line such that no two men are next to each other.
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