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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.

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter2: Systems Of Linear Equations
Section2.2: Direct Methods For Solving Linear Systems
Problem 3CEXP
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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.
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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