Arranging Colored balls

Question

There are 2 black balls, one red ball and one green ball. How many different linear arrangements can be generated by these balls ?

As per my understanding, the answer should be 4! * 3! (4 factorial because there are four spots that need to filled and 3 factorial because there are three distinct colors ). Is my approach correct ? or am i wrong ?

Answer 1

There are $\binom 42=6$ ways to choose positions for the two black balls.

For each of these choices, there are 2 ways to place the other two balls.

The total number of arrangements is $$N = 2\binom 42 =12$$

Answer 2

Treat their respective linear arrangement like letters in a four-digit word __ __ __ __ with the letters $b$, $g$, and $r$ for black green red respectively. There are $4*3*2*1$ spots for each letter but since there are 2 blacks we divide by $2!$ otherwise we would count $g,r,b1,b2$ and $g,r,b2,b1$ as different when they are indeed the same.

$4!/2!=12$

The answer wouldn't be $4!*3!$ because $4!$ is already counting the case of different colors (4 colors for the first spot, 3 for second...) you would instead divide by $2!$ (the number of ways the two blacks can be arranged)

Generally, questions like this can be solved with the formula $n!/(k1!k2!...kj!)$ the number of arrangments of a word (or line) consisting of $k1$ copies of object/letter 1,$k2$ copies of 2 and so on up to $kf$ copies of letter j, with n total letters/objects.

Answer 3

There is a rather similar approach that includes the 4! you were thinking of:

Let's say you've got 4 balls with 4 different colors. Then there are 4! ways to permute those.

Back to your case: Every single permutation has its permutation-buddy with the black balls switched. you can't make out any difference between those two permutations. Hence they are the same.

4!/2 = 12 possibilities

Edit: slower than @Leonard Euler