Discrete maths Combinatorics

Question

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

Comments

150?

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what is the right answer?

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25 @arch

Answer 1

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

Comments

Sir, why have you divided by 2 in case 1 ?

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the boxes are identical, the two different ways of selecting which box holds the first two letters and which one holds the second set of two letters will look the same. Hence, we need to divide the result by 2.

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Sir , So its 2! or just 2 ?

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Thank you so much for explaining.

Answer 2

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