The easiest and most elegant way to find a formula for this is to show you an equivalent problem. This is the case of indistinguishable objects in distinguishable boxes.
Rather than thinking that a Skittle is a particular color, let's have identical beads be placed in one of $5$ boxes, and depending on which box it is in determines its color of the Skittle.
Say we have boxes labeled "red", "green", "yellow", "orange", and "purple". If there were two red Skittles, that's the same as saying both beads were in the red box. If there was one green and one orange Skittle, then one bead was in the green box and the other in the orange box.
So we can count up the ways of putting the beads in boxes and it would be the same thing.
Now, instead of $5$ boxes, let's just split an area into $5$ regions using dividers. Let $|$ be a divider, like so:
$$
\square | \square | \square | \square | \square
$$
Now let's use stars for our beads, so the two red Skittles example could be represented with:
$$
\star \star | | | |
$$
Finally, we count up the ways of ordering the stars and stripes, which is the binomial coefficient $\binom{6}{2}$. In general, for $n$ Skittles and $k$ colors, the formula is $\binom{n+k-1}{k}$.
What is the probability that all three objects are the same color.
Your answer is correct.
Alternatively, we choose a ball. The probability that the second ball is the same color as the first is $\frac{5}{23}$. The probability that the third ball is the same color as the first two is $\frac{4}{22}$. Hence, the probability is
$$p = 1 \cdot \frac{5}{23} \cdot \frac{4}{22}$$
What is the probability that all three objects are different colors.
Your denominator is correct. We choose three of the four colors, then choose one object from each of the selected colors. Hence, the probability is
$$p = \frac{\dbinom{4}{3}\dbinom{6}{1}^3}{\dbinom{24}{3}}$$
Alternatively, we choose a ball. The probability that the second ball we select is of a different color than the first is $\frac{18}{23}$. The probability that the third ball we select is of a different color than the first two is $\frac{12}{22}$. Hence, the probability is
$$p = 1 \cdot \frac{18}{23} \cdot \frac{12}{22}$$
Best Answer
I apologize for my previous answer...I didn't read the problem carefully.
Let's focus on the number of ways we can pick 3 skittles such that they are all the different. Let's first look at the number of ways to pick a red, orange, and yellow skittle. There are $12$ red skittles, $16$ orange skittles, and $10$ yellow skittles so there are $12 \cdot 16 \cdot 10$ different ways of picking a red, orange, and yellow skittle.
To get the total number of ways to pick 3 skittles of different colors, just consider all the ways to pick $3$ colors out of $4$ and find the number of ways for each color combination. Doing this, we get a total of $$12 \cdot 16 \cdot 10 + 12\cdot 16 \cdot 18 + 16 \cdot 10 \cdot 18+12\cdot 10 \cdot 18$$ ways to pick $3$ skittles all of different colors.
Since there are ${12+16+10+18 \choose 3}$ ways to pick $3$ skittles in general, the probability is,
$$\frac{12 \cdot 16 \cdot 10 + 12\cdot 16 \cdot 18 + 16 \cdot 10 \cdot 18+12\cdot 10 \cdot 18}{{12+16+10+18 \choose 3}}$$