What's the probability that in three children 1 is a girl. It was asked that what is the probability that only one is a girl in a group of 3 children.
My sample space:- 
B B B 
B B G 
B G G 
G G G 
What my friend says:-
B B B 
B B G 
B G B 
G B B 
B G G 
G G B 
G B G 
G G G 
He says it will be like the when 3 coins are tossed. But I say that BBG and BGB are same. So there will be only 4 total outcomes.
Am I correct? Because if we see logically, then probability of one girl is 1/4. Because 1 girl is one girl, whether she is at left or right or middle.
And it is different from the 3 coins that we are tossing.

You're both correct on the sample space (in a sense); however, in your case, the outcomes are not equally likely.
Think about the order in which the children are born; there are $3$ ways to get two boys and one girl:

*

*First a boy is born, then another boy, then a girl

*First a boy, then a girl, then a boy

*First a girl, then a boy, then a boy

On the other hand, there is only one order in which three girls arrive, as another example. That is to say, "$2$ boys, $1$ girl" is not as likely as "$3$ girls". And this should make sense: think about your real life, and how often you've seen groups of three siblings, and how rare it's been an entire gender was excluded from them.
Having the sample space written in terms of equally likely events makes it significantly easier to calculate the relevant probabilities. Hence, even if each ordering of events results in the "same" outcome, in some sense, it's usually best to account for order when reasonable anyways.
This comes from a principle which says, if every event involved is equally likely,
$$\text{the probability $X$ happens} = \frac{\text{the number of ways $X$ can happen}}{\text{the number of possible events overall}}$$
Then one easily sees, for instance, the probability of "$2$ boys, $1$ girl" is $3/8$.
Since the events in your sample space are not equally likely, one has to use a different means of calculation.

An example taking your fallacy to the extreme is winning the lottery.
In your eyes, the sample space of events for winning a lottery has two events: "winning" and "losing". But if you try to calculate the probability in the erroneous method you have, you would conclude your probability of winning is $50\%$ (before we've even considered what the lottery is or how many people are in it, too!), which is obviously silly.
