There are 5 letters {a,b,c,d,e} are given.The boxes are identical.
Given that no box is empty.There are 2 possible cases for putting letter into boxes
a. 2, 2, 1
b. 3, 1, 1
Case A. 2, 2, 1
first Two letters out of the 5 can be selected in $5_{{C}_{2}}$ ways. Another 2 out of the remaining 3 can be selected in $3_{{C}_{2}}$ways and for the last latter, there is only one choice.
So.total Number of ways =$\frac{5_{{C}_{2}} \times 3_{{C}_{2}} \times 1}{2}$
=$15$
Case B. 3,1,1
Three letters out of the 5 can be selected in $5_{{C}_{3}}$ ways. For remaining boxes, there is one choice for each.
So,total number of ways = $5_{{C}_{3}} \times 1 \times 1 =10$
Total ways in which the 5 letters can be placed in 3 identical boxes =15+10 =25