Arranging a packet of skittles with repetitions of flavor

Question

A mini bag of skittles contains exactly 6 skittles. 3 red, 1 orange, 0 yellow, 0 green and 2 purple skittles. How many different ways are there to arrange the 6 skittles?

In order to solve this problem I was thinking of two different approaches however both give me different answers so I would like to know which is wrong and why. Thank you!

First approach: stars and bars

There would be 6 stars and 4 bars meaning I can arrange the skittles in 10C6 ways which evaluates to 210. Although the skittles are different colors the one's that are the same color are NOT distinct so this method would allow me to have an unordered set for the package of skittles.

Second approach: treat this as a permutation where I divide by any doubles. If the skittles were all unique they could be arranged in 6! ways however since some colors are repeated the total arrangements would be 6!/2!3!

Which approach is correct and why wouldn't these two approaches evaluate to the same answer?

Answer 1

Multinomial coefficient $\binom {6}{3,2,1} = \frac{6!}{3!\cdot 2!\cdot 1!} = 60$. This corresponds to your second approach.

I don't really understand how you are attempting to use stars and bars there.

Answer 2

Your second approach is correct because

$\quad (1)$: Any two skittles of the same color are indistinguishable

$\quad (2)$: In arrangements, ordering matters.

When one uses binomial numbers $($like $n$ choose $k$), ordering does not matter. Choosing a banana and an apple is the same as choosing an apple and a banana.

Answer 3

So in total you have $6$ skittles out of which $3$ are red, one is orange and $2$ are purple. And we need to arrange these skittles

So why don't you consider this problem as forming a $6$ letter word using $3$ R's , $1$ O and $2$ P's.

So the answer simply goes $$\frac {6!}{3!2!}$$