I have :
- $2$ apples
- $2$ oranges
- one pineapple
Apples,oranges are indistinguables
First of all I want to find the total number of possibilities so I think we are placing the pineapple at a fixed place and then permute the apples/oranges so : $$ \binom{4}{2} \cdot 5 $$
Now if I want to have all the possibilities and considering arrangement and its inverse (starting from the end) the same arrangement
For example : Apple-Orange-PineApple-Apple-Orange and Orange-Apple-PineApple-Orange-Apple
I was thinking we need to subtract the cases we have count several times So the only case we will have 2 same arrangement is when the pineapple is a the center so I was thinking all the possibilities will be:
The number of ways to create a "word" out of the letters $A,A,O,O,P$ is $$\frac{5!}{2!2!} = 30$$
Of these, only two of them are their own inverse: $AOPOA$ and $OAPAO$. This means the other $28$ have been double counted, so after dividing by $2$ and adding back the two whose inverse is the same we get $14+2=16$ ways.
This answer is also supported utilizing Burnside's formula, letting $|G|=2$:
$$\frac12 \left( 30 + 2\right) = 16$$ where the $2$ comes from the number of ways that the "word" stays the same if we flip it