Hi there. In this math post, I go over a math seating question that I helped out a student with during one of my tutoring sessions.
How many ways are there to seat 6 boys and 7 girls in a row of chairs so that none of the girls sit together?
You can assume that there are 13 seats/chairs available for the 6 boys and 7 girls. The condition of none of the girls sit together is the same as having girl, boy, girl, boy until the 7th girl. You have to have girls at the ends as there is 1 more girl than the number of boys.
Two Girls, One Boy Simple Case
When it comes to answering this question, the Fundamental Counting Principle does help. Consider a simple case where you have to seat 2 girls and 1 boy. For the left seat you have to place 1 of the 2 girls. This is 2 choices. Then you put the boy in the middle and place the last girl on the right. Diagram below with screenshot from TutorialsPoint Whiteboard.
There are 2 ways to seat 1 boy and 2 girls with the girls not next to each other.
Three Girls, Two Boys Case
What if we add 1 girl and 1 boy now. For the left spot, there are 3 girls to choose from. Then we have 2 boys to choose from.
Back to the original question. We want the number of ways to seat 6 boys and 7 girls so the girls are not next to each other.
For the left side, we have 7 choices from the girls. Then we seat 1 of 6 boys. The third seat from the left would be 1 girl out of the remaining six and continue on until girl number 7.
We would have 7! (7 factorial which is 7 x 6 x 5 x ... 1) for the girls multiplied by 6! (6 factorial) for the boys. Here is the calculation portion. (used Quicklatex).
There is a whopping 3 628 800 ways of seating 6 boys and 7 girls in a row of 13 chairs where no 2 girls are next to each other.
Thank you for reading.
