doubt about probability solution with $2$ girls and $2$ boys selection 
Q: A couple decides to have $4$ children. If they succeed in having $4$ children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly $2$ girls and $2$ boys?

Explanation of the solution:
Total number of ways of having $4$ kids is $1/2 \cdot 1/2 \cdot 1/2 \cdot 1/2 = 1/16$.
Total number of ways of having exactly $2$ girls and $2$ boys is:
First, count the $2$ girls and $2$ boys as $2$ girls glued together as one and $2$ boys glued together as one so there are $2!$ ways to move them.
Second, count the number of ways the $2$ girls can be moved within themselves $= 2!$
Third, count the number of ways the $2$ boys can be moved within themselves $= 2!$
So total is $2! + 2! + 2! = 6$
So the probability of then is $6/16$ or $3/8$.
What about GIRL, BOY, GIRL, BOY and BOY, GIRL, BOY, GIRL?? I am unable to understand where that was counted?
 A: It wasn't counted, the solution's explanation is erroneous (although the answer is correct). An easier way to compute this is to note that there are 2 choices for each of 4 children, so $2^4=16$ total arrangements. Out of these, $1$ corresponds to all $4$ being boys, $1$ corresponds to $4$ girls, and 4 each of $3$ boys $1$ girl, $3$ girls $1$ boy. Removing these 10 arrangements leaves only arrangements that are $2$ boys $2$ girls, so $16-10=6$. Thus the probability is $6/16$
A: The entire explanation is erroneous.
The number of possible sequences of boys and girls is $2^4 = 16$ since there are two choices for each of the four births.
The number of sequences with two boys and two girls is $$\binom{4}{2} = \frac{4!}{2!2!} = \frac{4 \cdot 3 \cdot 2 \cdot 1}{2 \cdot 1 \cdot 2 \cdot 1} = 2 \cdot 3 = 6$$ since exactly two of the four positions in the sequence of births must be reserved for girls.
The assumption that boys and girls are equally likely to be born means that the probability that the family consists of two boys and two girls can be found by dividing the number of favorable outcomes by the total number of outcomes.
Hence, the probability that a family with four children will consist of two boys and two girls if boys and girls are equally likely to be born is
$$\Pr(\text{two boys and two girls}) = \frac{\dbinom{4}{2}}{2^4} = \frac{6}{16} = \frac{3}{8}$$
Errors in the explanation
Note that the number of possible sequences of births must be an integer.  It cannot be $1/16$.
Children are born in a particular order.  The order of the girls cannot be switched, nor can the order of the boys, nor can the order of a block of boys and a block of girls.
What matters here is the positions of the boys and girls in the birth order.  We can count the favorable cases by choosing the positions of the girls in the birth order.  If there are two boys and two girls, then exactly two of the four positions must be filled by girls.  There are $\binom{4}{2} = 6$ ways to do this: ggbb, gbgb, gbbg, bggb, bgbg, bbgg.
The explanation seems to be an incorrect attempt to answer a different question, namely:
In how many ways can two boys and two girls be arranged in a line if the boys are adjacent and the girls are adjacent?
There are $2!$ ways to arrange a block of boys and a block of girls since either the boys precede the girls or the girls precede the boys.  For each such choice, the two boys can be arranged within their block in $2!$ ways and the two girls can be arranged within their block in $2!$ ways.  Hence, there are $2!2!2! = 2 \cdot 2 \cdot 2 = 8$ such arrangements.  To illustrate, suppose that the boys are Adam and Brendan and that the girls are Charlotte and Denise.  Then there are eight possible arrangements in which Adam is adjacent to Brendan and Charlotte is adjacent to Denise: ABCD, ABDC, BACD, BADC, CDAB, CDBA, DCAB, DCBA.
Addition should be used when two events are mutually exclusive, meaning that they cannot occur at the same time.  When two events can both occur, the number of ways they can both occur is found by multiplying the number of ways the first event can occur by the number of ways the second event can occur.
A: Solution :
We know,

              total number of outcomes that fullfill the requirement 
 Probabilty = ------------------------------------------------------
                    total number of all possible  outcomes


In the case we want probability of  girl being born ;

among all possible outcomes of (girl,boy) our desired outcome is birth of girl.

so total number of outcomes that fullfill the requirement = 1 

while  total number of outcomes = 2

    So ; Probability of Girl Birth P(G) = 1/2 ; 

    Similarly ;Probability of Boy Birth  P(B) = 1/2

Now consider A case where
                first birth is girl ;
                Second Birth is a girl;
                Third birth is a boy;
                Fourth birth is a boy;

let the above case be represented as GGBB.
Proabability of GGBB is given by :

P(GGBB) = P(G) x P(G) x  P(B) x P(B)
    = 1/16


GGBB is just one of many possible outcomes where 2 Girls and 2 Boys are born in 4 Births. 
there are other combinations of births that can give  2 Girls and 2 Boys
including GGBB. they are :

GGBB
BBGG
BGBG
BGGB
GBBG
GBGB

where each have probability of 1/16.


So probability of (2Girls and 2Boys)    = Sum of probability that result in birth of two boys and two girls
                    = P(GGBB)+P(BBGG)+P(BGBG)+P(BGGB)+P(GBBG)+P(GBGB)
                    = 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16
                    = 6/16

So basicallly probability of (2Girls and 2Boys) ie. P(2G2B) = total number of ways 2 girls and 2 boys can be born * probability of any one case where 2 boys and 2 girls is born
                        = 6 * P(GGBB)

So if we can find the total no. of ways  2 girls and 2 boys can be born then we can find P(Exact 2G2B) 
without having to list all possibilities of ways  2 girls and 2 boys could be born.

How do you find total no. of ways  2 girls and 2 boys can be born?

We use the formula nCk = n!/(k!(n-k)!) .

Understanding nCk before using it :

nCk gives the total no. of ways , we can arrange total of n objects into  combinations of k ,
where the ordering of combinations made by  selected objects doesnt matter .

meaning if we have 5 Cars - A B C D E 

5C3 gives 10. 
meaning if we select and order the cars in groups of 3:

ABC
ABD
ABE
ACD
ACE
ADE
BCD
BCE
BDE
CDE

total = 10 

And the ordering of combinations made by  selected objects doesnt matter means we didnt include  variations of combinations .
eg variation of first combination ABC are -  BCA  ,ACB, CAB so on theyre considered same as ABC.


so nCk gives the total number of combinations of k possible;from total objects  n . 

Now How does this help to find  total no. of ways 2 girls and 2 boys can be born .?

Consider  " 1 2 3 4 " as 1st-birth 2nd-birth 3rd-Birth and 4th-birth.

let us use 4C2 on the above ; meaning 4C2 gives total no. of ways the objects 1 2 3 4 can be arranged in pairs of 2.

ie.
  
 1 2
 1 3
 1 4
 2 3
 2 4
 3 4

so 4C2 = 6

Also , Indirectly the above combinations i wrote gives u all the positions u could put giving birth to boy ; 
1 2 or 1 3  or 1 4 or 2 3 or 2 4 or 3 4
so automatically the remaining position becomes giving birth to girl.

So combinations could be - 
BBGG
BGBG
BGGB
GBBG
GBGB
GGBB

or vice versa  these position could be used to put giving birth to a girl.
so automatically the remaining position becomes girl.

So indirectly 4C2 gave us total no. of ways 2Girl 2Boys could be born. 

so the way of solving the question  without having to list all possibilities of ways  2 girls and 2 boys could be born is = 4C2 * P(GGBB)
                                                    
                                                              = 6 * 1/16
                                                              = 6/16 [Soln]


