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How many ways can the letters $\{a,b,c,d,e\}$ be placed into $3$ identical boxes such that no box is empty?

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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

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you can also solve by assuming that these boxes are different ......and way of distributing these 5 balls in 3 different boxes is 150......since the boxes are same divide it by 3!.......ans is 25

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