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?

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]


