<s>[INST] Probability of drawing objects: Combinations vs Permutations 
A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random without replacement the probability of getting exactly one red ball is

Attempt 1
$$
P(\text{exactly 1 red ball})=\frac{^5C_1 \cdot ^3C_2}{^8C_3}=\frac{5 \cdot 3}{\frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1}}=\frac{5 \cdot 3 \cdot 3 \cdot 2}{8 \cdot 7 \cdot 6}=\frac{15}{56}
$$
Attempt 2
\begin{align*}
P(\text{exactly 1 red ball}) & = P(RBB)+P(BRB)+P(BBR)\\
& =\frac{5 \cdot 3 \cdot 2}{8 \cdot 7 \cdot 6}+\frac{3 \cdot 5 \cdot 2}{8 \cdot 7 \cdot 6}+\frac{3 \cdot 2 \cdot 5}{8 \cdot 7 \cdot 6}\\
& =3 \cdot \frac{5 \cdot 3 \cdot 2}{8 \cdot 7 \cdot 6}\\
& =\frac{15}{56}
\end{align*}
Though I get same result in both methods, I think first method uses combinations(order does not matter) and second uses permutation(order matters), how can I differentiate the two ?
 [/INST] In Attempt 1 the underlying assumption is that you have a random mechanism selecting a $3$-element subset from a given $8$-element set, whereby all $3$-element subsets are equiprobable. You then count the number of favorable $3$-element subsets.
In Attempt 2 the underlying assumption is that you have a random mechanism selecting an element from a given finite set, whereby all elements are equiprobable, and you apply this mechanism three times. Finally you compute the probability of the desired outcome. 
The reason that you obtain the same result in both cases is the following: Both mechanisms select (in one, resp., in three steps) a $3$-element subset of the given $8$-element set, uniformly over all ${8\choose3}$ such subsets.The first mechanism does so by its specification, and the second "by symmetry".
</s>