ember hornet
Let's say I have the following balls:
RRRWW
(R: red, W: white)
Then there are 5 choose 3 combinations, since:
5 * 4 * 3 different ways to arrange the red balls and then divide by 3! since the red balls are indistinguishable. Then automatically you take care of the white balls.
But what if you have:
RRRWWGG
How many different ways can you arrange this now? (G: green)
shut nightBOT
ember hornet
RRRWW: 5! / 3! / 2!
RRRWWGG: 7! / 3! / 2! / 2!
Ooh wait I see that actually makes sense
So let's look at RRRWW, is my following logic correct?:
Lets pretend first that every ball is distinguishable, then there are 5! ways to arrange. But now you have 3 red balls, which can be arranged in 3! ways, so you divide by 3! and the same for the white balls
My previous logic was:
RRRWW is the same as placing 3 red balls on 5 empty squares.
So 5 * 4 * 3 possibilities to place these red balls, and rearranging these 3 red balls should not be counted, so 5 * 4 * 3 / (3!)
But that logic could not be extended to multiple colors, but with your logic it can