Finding the number of arrangements

[hms.tech]

Homework Statement

In how many ways can 3 red and 3 blue balls be arranged if no two balls of the same colour are next to each other ?

Homework Equations

...

The Attempt at a Solution

Here is what i did (Which is wrong)

6 x 3 x 2 x 2 x 1 x 1

I did this since there are 6 possibilities for the first spot.
there are 3 possibilities for the second spot
there are only 2 possibilities for the 3rd spot and so on ...

The correct answer is 144 (which is essentially double my answer)

 

[HallsofIvy]

The crucial question here, and you don't answer it, is "are the same colored balls identical or distinguishable?" Since there are only two kinds of balls and balls of the same color cannot be next to each other, you must alternate the balls. If the same color balls are identical, there are only two arrangemennts- BRBRBR or RBRBRB. If all balls are distingushable we can
First choose a blue or red ball to go first: there at 2 ways to do that.
Having chosen, say, a blue ball, there are 3 to choose as the first ball, then 3 red balls to choose second, 2 blue for third, two red for fourth, a blue for fifth and a blue for second. If the first ball was red, we have the same situation, reversing red and blue.

There are 2(3)(3)(2)(2)(1)(1)= 72 ways to do that. I think you are correct and "144" is incorrect.

 

[hms.tech]

Lets put our method to test, what if the number of red balls and blue balls were not equal .
There are 5 red balls (distinguishable) and 3 blue balls (distinguishable) , find the number of possible permutations in which all blue balls are not next to each other .

I would speculate :

for the first spot : 8
(assuming the first was blue) second spot : 5
third spot : 2
fourth : 4
fifth : 1
sixth : 3
seventh : 2
last : 1

That's total : 1920

Agreed ?

 

[CAF123]

Lets put our method to test, what if the number of red balls and blue balls were not equal .
There are 5 red balls (distinguishable) and 3 blue balls (distinguishable) , find the number of possible permutations in which all blue balls are not next to each other .

I would speculate :

for the first spot : 8
(assuming the first was blue) second spot : 5
third spot : 2
fourth : 4
fifth : 1
sixth : 3
seventh : 2
last : 1

That's total : 1920

Agreed ?

I think this assumes the ordering BRBRBRRR because at the third spot you only assign two possible options (which I think you mean the two remaining blue balls). But the third spot can also be red, ie something like BRRRBRRB is acceptable).

 

[hms.tech]

I think this assumes the ordering BRBRBRRR because at the third spot you only assign two possible options (which I think you mean the two remaining blue balls). But the third spot can also be red, ie something like BRRRBRRB is acceptable).

*amazed* You are right.

So how would we proceed in this case , do u have any leads ?

 

[Ray Vickson]

*amazed* You are right.

So how would we proceed in this case , do u have any leads ?

There are 5! permutations of the 5 red balls. There are 6 places where we can put the 3 blue balls (before each of the reds and after the last red). In how many permutations can you put 3 blue balls in 6 places?

BTW: that method yields 144 arrangements in your original problem.

 

[rcgldr]

There are 3! = 6 ways to permute the 3 red balls, and there are 3! = 6 ways to permute the 3 blue balls. There are two permuations for the 6 balls together, RBRBRB or BRBRBR, so that would seem to produce a total of 6 x 6 x 2 = 72 possible permutations, which is the same answer as in the original post. I'm not sure if we're missing something here.

 

[haruspex]

BTW: that method yields 144 arrangements in your original problem.

Except, the original problem also forbade adjacent red balls, which halves the combinations.

 

[Ray Vickson]

There are 3! = 6 ways to permute the 3 red balls, and there are 3! = 6 ways to permute the 3 blue balls. There are two permuations for the 6 balls together, RBRBRB or BRBRBR, so that would seem to produce a total of 6 x 6 x 2 = 72 possible permutations, which is the same answer as in the original post. I'm not sure if we're missing something here.

Sorry: you are right, of course. The method I described was for reds being allowed to touch but not blues. If neither can touch themselves, that cuts the possibilities in half.

 

[hms.tech]

Sorry: you are right, of course. The method I described was for reds being allowed to touch but not blues. If neither can touch themselves, that cuts the possibilities in half.

Which problem are u addressing :
1. 3 Reds and 3 Blues
2. 6 red and 3 blues

If the FIRST one then here is my question to u :
How can two red balls touch each other for it is given that the blue balls must be apart from each other . If any red ball was to touch another red ball, the two of the blue balls would be bound to touch .

 

[Ray Vickson]

Which problem are u addressing :
1. 3 Reds and 3 Blues
2. 6 red and 3 blues

If the FIRST one then here is my question to u :
How can two red balls touch each other for it is given that the blue balls must be apart from each other . If any red ball was to touch another red ball, the two of the blue balls would be bound to touch .

I am answering the question posed before, where there are more red than blue balls, and reds are allowed to touch each other but blues are not.