Probability on randomly selecting 3 balls from a bowl of 6 white and 5 black balls This is an example from a book on probability. 
If 3 balls are “randomly drawn” from a bowl containing 6 white and 5 black balls,
what is the probability that one of the balls is white and the other two black?
Let $A$ be the event where the order of drawing balls mathers, $P(A) =\dfrac{6\cdot 5\cdot 4 \cdot 3}{11\cdot 10\cdot 9}$.
Let $B$ be the event where the order of drawing balls does not mather, $P(B)=\dfrac{\binom{6}{1} \binom{5}{2}}{\binom{11}{3}}$.
Both leads to the same probability, as $P(A) = P(B) \cdot \dfrac{3!}{3!}$.
My question is on the counting part  of my above solution.
I have deemed the white balls as distinguishable among themselves, and so do the black balls.
For example, there are 6 white balls, each labelled from 1 to 6, and the black balls labelled from 1 to 5.
For example, there are $\dbinom{5}{2}$ ways to choose 2 black balls from the 5 black balls where order does not mather.
If i would to list all the possible outcomes, it would be $\bigg\{\{b_1,b_2\} , \{b_1,b_3\}, \{b_1,b_4\}, \{b_1,b_5\}, \{b_2,b_3\}, \{b_2,b_4\}, \{b_2,b_5\}, \{b_3,b_4\}, \{b_3, b_5\}, \{b_4,b_5\}\bigg\}$
So i have been counting in this manner since my first course in probability, which i have "label" or distinguish the balls of the same colour among themselves.
But what is the reason that this is the coorect way to count in the first place?
If i were to replace the white balls as men, and the black balls as women, then it is intuitive that the each man is different, and each woman is different. But this is a different scenario.
 A: It is okay to label and list all possible outcomes, but it becomes extremely tedious and even near impossible when you have large sets. When you label the balls you actually change the nature of the variable and thus requires a different solution route. Yet, you seem to understand this as both of your approaches were sufficient for this problem.
So to directly answer your question, there is no need to label the balls in these kinds of problems. In fact, I recommend using the approach you did for $P(B)$ as this will be the most useful in difficult counting/probability problems and in this particular case order does not matter.
A: It's normal to label the items and maybe assume an order. Otherwise you couldn't easily count things (by product rules etc.). In the end you just discard the labels (and the order, if need be). It doesn't matter to the probability whether we throw 2 marked dice or two unmarked ones, right? But the marked ones allow us to see that there really are 2 different ways to throw $(1,2)$, and so the probability to throw sum $3$ is twice that of sum $2$. Labelling shows us the real underlying truth. 
Using labelling we show that there are $5 \choose 2$ ways to pick 2 objects out of 5, without order and no laying back. We can us this as a lemma in other more complicated problems. 
So for the ball problem we have to pick one white ball and two blacks and the order per colour does not matter. So we have ${6 \choose 1}$ and ${5 \choose 2}$ respectively, and the total number of options without order is their product. The lemma provides us with ${11 \choose 3}$ total options. Hence the quotient.
A: 
If i were to replace the white balls as men, and the black balls as women, then it is intuitive that the each man is different, and each woman is different. But this is a different scenario.

No, it is basically the same situtation.   You are selecting from physically distinct objects which may share an identifying property.
Being unable to visually distinguish balls of the same colour does not mean that they are somehow the same ball.
When measuring the probability for an event, it is not sufficient to merely count the distinguishable outcomes in the sample space; you have to weigh the probability for each of those outcomes. 
( When buying a ticket in a lottery there are two distinct outcomes: either you win xor you lose.  Is the probability that you win usually equal to $1/2$?)
