# Very tricky line of people problem

• aeronautical
In summary: A1, the third ball in A2, and so on.The problem is that you're trying to solve for the number of ways in which you can put a certain number of balls into a certain number of slots, when in fact you need to solve for the number of ways in which you can place a certain number of balls into a certain number of slots, given that the ordering of one of the balls is known.
aeronautical

## Homework Statement

Imagine that there are two lines of people standing by two different cashier spots (cashier 1 and cashier 2) in a department store. A_{1} stands first in line by cashier #1,and is followed accordingly by, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}. In the same manner B_{1} stands first in line by cashier #2, and is followed accordingly by, B_{2}, B_{3}, B_{4}, and B_{5}. Suddenly both Cashier #1, Cashier #2 close their cashier spots and hence, a new line has to be formed all the people in both lines by a completely new Cashier spot, under the condition that the order, A_{1}..A_{6} is maintained just as the order B_{1}..B_{5} is maintained. Furthermore, will A_{3} and B_{2} whom haven't met each other for a long time (and need to talk), stand next to each other in the new formed line. In how many ways may the new line look like?

## Homework Equations

I have tried to understand this problem, how can they stand next to one another if the order has to be maintained. Can somebody shed some light upon this problem...please show all steps... there must be a more efficient way then the way I present below :(

## The Attempt at a Solution

If one considers different orders, you can end up with SO many different possibilities... () marks B2 and A3 they need to be next to each other

Option 1: A1 B1 A2 (B2 A3) B3 A4 B4 A5 B5 A6
Option 2: A1 A2 B1 (B2 A3) B3 B4 B5 A4 A5 A6
Option 3: A1 A2 B1 (B2 A3) A4 A5 A6 B3 B4 B5
Option 4: A1 A2 B1 (B2 A3) B3 A4 B4 B5 A5 A6
Option 5: A1 A2 B1 (B2 A3) B3 A4 A5 B4 B5 A6
Option 6: A1 A2 B1 (B2 A3) B3 A4 A5 A6 B4 B5
Option 7: A1 A2 B1 (B2 A3) B3 B4 A4 A5 A6 B5
Option 8: A1 A2 B1 (B2 A3) B3 B4 A4 B5 A5 A6
Option 9: A1 B1 A2 (A3 B2) B3 A4 B4 A5 B5 A6
Option 10: A1 A2 B1 (A3 B2) B3 B4 B5 A4 A5 A6
Option 11: A1 A2 B1 (A3 B2) A4 A5 A6 B3 B4 B5
Option 12: A1 A2 B1 (A3 B2) B3 A4 B4 B5 A5 A6
Option 13: A1 A2 B1 (A3 B2) B3 A4 A5 B4 B5 A6
Option 14: A1 A2 B1 (A3 B2) B3 A4 A5 A6 B4 B5
Option 15: A1 A2 B1 (A3 B2) B3 B4 A4 A5 A6 B5
Option 16: A1 A2 B1 (A3 B2) B3 B4 A4 B5 A5 A6

...GAH

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How many ways are there to place k people in n slots? You'll need to do two such calculations--one for A and one for B. Orderings for A are independent of orderings for B. Use the constraint {...A2B3...} U {...B3A2...} at the end to reduce the number of possible orderings (here "U" means union).

thanks for shedding some light, the problem is that i can't figure it out. i know that there is a way, i just never got into probability. I don't know how to solve this

Here's a hint: suppose you've found the number of ways you can put 6 balls into 11 slots (orderings of A). One such ordering might be

(A1 X X A2 X A3 X A4 A5 X A6)

Each X can hold a person from line B. How many ways are there to place {B} in this configuration? Just one:

(A1 B1 B2 A2 B3 A3 B4 A4 A5 B5 A6)

This goes for all orderings of A (you could start with B and get the same result). The next thing you'll need to do is constrain the number of possible orderings by the constraint.

bigplanet401 said:
Here's a hint: suppose you've found the number of ways you can put 6 balls into 11 slots (orderings of A). One such ordering might be

(A1 X X A2 X A3 X A4 A5 X A6)

Each X can hold a person from line B. How many ways are there to place {B} in this configuration? Just one:

(A1 B1 B2 A2 B3 A3 B4 A4 A5 B5 A6)

This goes for all orderings of A (you could start with B and get the same result). The next thing you'll need to do is constrain the number of possible orderings by the constraint.

Thanks...well I am just confused why you say 'just one'. Cuz If i consider the second case it is 2. Cause I can put 5 balls into 11 slots...only two ways. And when i consider the union I know that the possibilities will be even less...how can i account for the number of possibilities is it only 2 possibilities since 2*1=2?

The ordering of B is completely determined by the ordering of A:

(A1 X X A2 X A3 X A4 A5 X A6)

means {B} must fill in the X's in the following way:

(A1 B1 B2 A2 B3 A3 B4 A4 A5 B5 A6)

It can be no other way--this is the only possible ordering of the B's given the ordering of A above. You can put the A's into line 11C6 possible ways, where 11C6 means "11 choose 6". For each ordering of the A's, you get one, and only one, possible ordering for the B's.

I can follow you on what you wrote. The ordering of Bs are determined by As. So you mean to say that the number of possibilities are 11-6=5 possibities and the constraint brings that down to 5 minus 2 = 3 possibilities?

11C6 is a binomial coefficient: have a http://en.wikipedia.org/wiki/Binomial_coefficient" .

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bigplanet401 said:
11C6 is a binomial coefficient: have a http://en.wikipedia.org/wiki/Binomial_coefficient" .

Thanks...I have started to look at this problem from the beginning...and read some. Isn't the following a solution?

My answer should be the product of of three numbers:

N1 = the number of ways the people before the A3 B2 pairing can be arranged

N2 = the number of ways the people after the A3 B2 pairing can be arranged

2 = the number of ways A3 and B2 can be arranged.

I conclude that N1 is 3; just by looking at the listed possibilities above or 3-choose-1.(3 = 3! / (1! (3-1)!) = (3*2*1)/(1*(2*1))=6/2=3
2)

So the problem is all about solving for N2.

Well, there are a total of 6 people after the A3 B2 pairing. 3 of them are As and 3 are Bs. If I know which 3 are Bs, I should be able to figure EVERYTHING else out about the order of the 6 people. So N2 should be equal to 6-choose-3!

(6 = 6! / (3! (3-6)!) = (6*5*4*3*2*1)/((3*2*1)*(3*2*1))=120/6=20
3)Isn't the answer to this problem (the number of possibilities)

3 * 2 * 6-choose-3 = 3 * 2 * 20 = 6 * 20 = 120

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## What is the "Very tricky line of people problem"?

The "Very tricky line of people problem" is a mathematical problem that involves arranging a group of people in a line according to certain rules or criteria.

## What makes the "Very tricky line of people problem" difficult?

The "Very tricky line of people problem" can be difficult because it often involves complex mathematical concepts such as permutations, combinations, and probability.

## What are some strategies for solving the "Very tricky line of people problem"?

One strategy for solving the "Very tricky line of people problem" is to break it down into smaller, more manageable problems. Another strategy is to use logic and reasoning to eliminate impossible or unlikely scenarios.

## Is there a specific formula for solving the "Very tricky line of people problem"?

No, there is not a specific formula for solving the "Very tricky line of people problem." It often requires creative thinking and problem-solving skills.

## Why is the "Very tricky line of people problem" important?

The "Very tricky line of people problem" can help improve critical thinking and problem-solving skills, which are essential in many fields of science, engineering, and mathematics.

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