Insert 3 distinct rings into 2 fingers and insert 2 fingers into 3 distinct rings. Here are the problems I am unable to grasp.
A- Insert $3$ distinct rings into $2$ fingers.
and
B- insert $2$ fingers into $3$ distinct rings.
Both questions seem similar but 1st one has $2×2×2=8$ ways from multiplication principle and 2nd one has $3×3= 9$ ways from multiplication principle.
When I try to figure out number of ways by manual counting I get only 6 ways for both questions
For question (A) I get-$(r1,f1),(r1,f2), (R2,f1), (R2,f2), (r3,f1), (r3,f2)$
For question (B) I get- $(f1,r1),(f1,R2), (f1,r3), (f2,r1), (f2,R2) , (f2,r3)$.
Why my answers are different i.e. only 6 ways and not 8 or 9 ways? Can any one provide correct manual counting?
 A: I think your part A is wrong , so i need to write this answer.
PART A-)
This question is a type of flags into poles where the order of flags relevant.As you know , the order of rings msut be different in a finger , because one of them will be at bottom , the others are upper. For example , if we have $3$ rings  ,then we can insert them into a finger in $3!$ ways.
Now , we should handle that how many rings will be inserted by each fingers.For this , we will use stars and bars method. According to the method , we can distribute $3$ identical objects into $2$ distinct boxes (i.e fingers) by $$\binom{3+2-1}{3}=4$$ For example ,$*|**$ is one of the possible arrangement where stars represent the palces for rings .Moreover the left of the bar can be thought as the first finger and the right of the bar can be thought as the second finger. Now , lets place our $3$ distinct rings into the possible places which denoted by "stars".
We can place them $3!$ ways. Then , the answer is $$3!\binom{3+2-1}{3}=24$$
