Let's say, I have 4 yellow and 5 blue balls. How do I calculate in how many different orders I can place them? And what if I also have 3 red balls?
Combinatorics – Different Ways to Sort Balls of Two Different Colors
combinatorics
combinatorics
Let's say, I have 4 yellow and 5 blue balls. How do I calculate in how many different orders I can place them? And what if I also have 3 red balls?
Best Answer
This is a standard problem involving the combinations of sets, though perhaps not very obvious intuitively.
Firstly consider the number of ways you can rearrange the entire set of balls, counting each ball as indepndent (effectively ignoring colours for now). This is simply $(4 + 5)! = 9!$, since the 1st ball can be any of the $9$, the 2nd can be any of the remaining $8$, and so on.
Then we calculate how many different ways the yellow balls can be arranged within themselves, since for the purpose of this problem they are considered equivalent. The number of combinations is of course $4!$; similarly, for the blue balls the number is $5!$.
Hence, overall we find:
$$\text{total arrangements} = \frac{\text{arrangements of all balls}}{\text{arrangements of yellow balls} \times \text{arrangements of blue balls}}$$
Therefore in our case we have:
$$\text{total arrangements} = \frac{9!}{5! \times 4!} = 126$$
I'm sure you can see how this can be easily extended if we also have 3 red balls too. (Hint: the total changes and we have another multiple of identical arrangements to account for.)