# Counting problem -- Lining up colored marbles....

• member 587159
Your mistake was in the second line, where you wrote 72 - (8 nCr 1) = 64. The correct calculation should be 72 - (8 nCr 2) = 56.
member 587159

## Homework Statement

I have 3 yellow and 7 blue marbles. I put those randomly in a line. What is the probability that no yellow marbles are next to each other?

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## The Attempt at a Solution

Total amount of opportunities to put those marbles next to each other: 120
Total amount of opportunities to put 3 marbles next to each other: 8
Total amount of opportunities to put 2 marbles next to each other:

(9 nCr 1)*(8 nCr 1) = 72
72 - (8 nCr 1) = 64

1 - (64+8)/120 = 2/5

Can someone verify whether I'm correct? Or at least tell me where I am wrong?

I agree that the total number of color patterns is 120 = 10!/(7!*3!).
When I calculate the number of arrangements for no double yellow, I did it with two scenarios:
1) Turn all three yellow marbles into yellow-blue pairs. That leaves three pairs and four extra blue.
2) Take one yellow and put it at the end. Then pair the remaining 2 yellows into 2 yellow-blue pairs. That leaves 5 extra blues.
The total of those two numbers gives you the total number of arrangements that keep the yellows separated.

Did you get 56/120? I seem to have made a mistake.

.Scott said:
I agree that the total number of color patterns is 120 = 10!/(7!*3!).
When I calculate the number of arrangements for no double yellow, I did it with two scenarios:
1) Turn all three yellow marbles into yellow-blue pairs. That leaves three pairs and four extra blue.
2) Take one yellow and put it at the end. Then pair the remaining 2 yellows into 2 yellow-blue pairs. That leaves 5 extra blues.
The total of those two numbers gives you the total number of arrangements that keep the yellows separated.

Small refinement to your method: create one extra blue as well as coupling a blue with each yellow. In the blue-yellow pairs, have the blue on the right always. So there will always be a blue at the right end of the line, which you throw away to get back to 7. So the answer is the number of ways f choosing 3 YB pairs from 3+(7-3+1) things.

Math_QED said:
(9 nCr 1)*(8 nCr 1) = 72
72 - (8 nCr 1) = 64
Why 9 and 8? What does the second line do?

Math_QED said:
Did you get 56/120? I seem to have made a mistake.

Your 56/120 is correct. Here is how I looked at the problem: line up the 7 blues, with spaces between them. You have 8 spaces altogether (one to the left of all the blues, 6 between the blues, and one to the right of all the blues). Now you want to put three yellows into those 8 spaces, with either 0 or 1 per space. The number of ways of doing that is the number of ways to pick 3 spaces from 8, which is C(8,3) = 56. The number of ways of arranging the 10 marbles (with no restrictions) is the number of ways of choosing the 3 positions in which the yellows will go, so is C(10,3) = 120. Therefore, the probability you want is 56/120 = 7/15.

mfb and member 587159
Ray Vickson said:
Your 56/120 is correct. Here is how I looked at the problem: line up the 7 blues, with spaces between them. You have 8 spaces altogether (one to the left of all the blues, 6 between the blues, and one to the right of all the blues). Now you want to put three yellows into those 8 spaces, with either 0 or 1 per space. The number of ways of doing that is the number of ways to pick 3 spaces from 8, which is C(8,3) = 56. The number of ways of arranging the 10 marbles (with no restrictions) is the number of ways of choosing the 3 positions in which the yellows will go, so is C(10,3) = 120. Therefore, the probability you want is 56/120 = 7/15.

Wow. Nice method. I greatly appreciate your effort and clear explanation.

Math_QED said:
Did you get 56/120? I seem to have made a mistake.
yes

member 587159

## What is the counting problem?

The counting problem refers to the challenge of determining the number of ways a given set of objects can be arranged or ordered.

## What is the "lining up colored marbles" counting problem?

The "lining up colored marbles" counting problem specifically involves determining the number of ways a set of differently colored marbles can be arranged in a line.

## Why is the counting problem important?

The counting problem is important in various fields such as mathematics, computer science, and statistics. It allows us to find the total number of possible outcomes in a given situation and make predictions or solve problems based on those outcomes.

## What is the formula for solving the counting problem?

There are various formulas that can be used to solve the counting problem, depending on the specific scenario. One common formula is the factorial formula, which is n! = n x (n-1) x (n-2) x ... x 2 x 1, where n represents the number of objects being arranged.

## Are there any real-life applications of the counting problem?

Yes, the counting problem has many real-life applications, such as in probability and statistics, genetics, and permutations and combinations in mathematics. It can also be used to solve problems related to arranging objects in a specific order, such as seating arrangements at events or scheduling tasks for a project.

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