Finding number of elements common in two set of permutation? How to find the number of common terms in two sets of permutation? 
I need to find the number of 4-letter words that can be formed out of the letters {a,b,c,d} with the conditions, 


*

*Letter 'b' appear before c and d


Total number of words from letters {a,b,c,d} = 4! = 24
Let {A} - set A containing words where 'b' appear before 'c'(but 'b' may or may not appear before 'd')
 {A} =  {abcd,abdc,adbc,bcda,bcad,bdca,bdac,badc,bacd,dabc,dbca,dbac} = 4!/2! = 12
Let {B} - set B containing words where 'b' appear before 'd'(but 'b' may or may not appear before 'c')
 {B} =  {abcd,abdc,acbd,bcda,bcad,bdca,bdac,badc,bacd,cabd,cbad,cbda} = 4!/2! = 12
I need to find out the total number of words which appear in both set A and B. That is, the words in which b appear both before c and d. 
 A intersect B = {abcd,abdc,bcda,bcad,bdca,bdac,badc,bacd} = 8
 How to find the number of words in A intersect B ?
 A: Symmetry lets us handle all these questions.
To answer the original one:  Consider any permutation of your letters.  Within it, $\{b,c,d\}$ occur in some order and all orders are equally likely.  Of the six  possible permutations of $\{b,c,d\}$, $b$ comes first in two.  Hence $b$ is first in a third of the total permutations, so the answer is $$\frac {4!}{3}=8$$
Of course, as $4$ is so small we could simply have enumerated the cases.  They are $$\{bacd,\,badc,\,bcad,\,bdac,\,bcda,\,bdca,\,abcd,\,abdc\}$$
But the symmetry argument works for larger collections.
Counting $A\cap B$ is equivalent to the original question, so the answer is $8$.
A: I do not know if there is a direct way to compute the interesction, but the following works: A word, where both $c$ and $d$ appear after $b$ must start with either $ab$ or $b$, let's count both groups:


*

*We have $2! = 2$ words starting with $ab$, as only 2 letters are left to distribute

*We have $3! = 6$ words starting with $b$, as only 3 letters are left to distribute


So alltogether there are $2 + 6 = 8$ words in $A \cap B$.
