Question 1175991: There are 5 boys and 4 girls forming a line to pay for the ticket to watch Frozen 2. In how many ways can they be arranged if 3 particular persons must stand beside each other?
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
Let's say these 5 boys and 4 girls are students of a small classroom. Let's also say that the teacher is chaperoning them.
The 5+4 = 9 students are standing in line, and let's say that 3 adjacent students (standing next to each other) leave the line to get popcorn. The teacher can step in for those 3 students as a substitution. Wherever the teacher is, the three students will replace him/her when the students get back from buying popcorn.
So the 9 people in line drops to 9-3 = 6 people after those 3 students leave. But then the teacher steps in to bump the count to 6+1 = 7 people.
Those 7 people arrange in 7! = 7*6*5*4*3*2*1 = 5040 ways.
The exclamation mark indicates factorial. We count down 7,6,5,... all the way to 1 multiplying along the way.
There are 5040 ways to arrange the group of students+teacher. Order matters. Of any of those 5040 permutations, there are 3! = 3*2*1 = 6 ways to arrange the trio of students who left the line.
Consider those three students to have codenames of A,B,C
Those 6 arrangements are such
A,B,C
A,C,B
B,A,C
B,C,A
C,A,B
C,B,A
This means that we'll have 6*5040 = 30,240 different ways to form the line such that those 3 students must stand next to one another. At this point, the teacher has been removed from the line.
|
|
|
