It is a classic problem on bars and stars method.


You need to find the number of all NON-negative integer solutions of the equation

x1 + x2 + x3 = 4     (1)


Imagine 4 letters on the table, placed in one straight line with small gaps between them.


Imagine you have 4 numbered bars, of which you place the first bar (bar N1) before the first letter and the last bar (bar N4) 
after the last letter in the row.


Let  , ,   be some solution to the given equation.


You then place bar N2 after -th letter in the gap in the row of letters; 

then you count next  letters  in the row of letters after bar N2 and place bar N3 in the gap there;

The last  letters are between bars #3 and #4.


At this moment, all 4 letters are divided in 3 groups between bars (1-2), (2-3), and (3-4).


Notice that if some  is zero, then the corresponding bars go to the common respective gap.


So, having the solution to equation (1) in non-negative integer numbers, you place 2 bars, N2 and N3, in
their corresponding positions in gaps in the row of letters. 


Vise versa, if you place 2 bars B2 and B3 in gaps in the row of 4 letters, you divide letters in 3 groups, 
and the numbers of marbles in each group form the solution to equation (1).


Thus, there is one-to-one correspondence between the set of solutions to equation (1) in non-negative integer numbers, 

from one side, and all different possible placings of 2 bars #2 and #3 in 2 gaps in the row between 4 letters.


Thus we have 4 + 2 = 6 entities, 4 letters and 2 movable bars #2 and #3; 4 letters are indistinguishable and 2 movable bars 
are indistinguishable, too.


The number of all possible indistinguishable arrangements of 6 items of two types with 4 indistinguishable of one type 

(letters) and 2 indistinguishable of the other type (bars) is  

     =  = 15.


Hence,  the number of all possible solutions to equation (1) in non-nrgative integer numbers is 15.   

ANSWER.  The number of different placings of 4 letters in 3 mailboxes is  15.