Kenneth Rosen Edition 6th Exercise 5.5 Example 10 (Page No. 377)

Question

How many ways are there to put four different employees into three indistinguishable offices when each office can contain any number of employees?

Comments

Answer will be 14 ways

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can you explain please?

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Can an employee be present in more than one office at a time? Since an office can also mean a department and an employee can be included in more than one department at a same time. Question hasn't clearly defined the mapping/relationship between employees and offices. Similarly, there could be cases when one or more office has no(zero) employees in it because the last line of the question allows to do so.

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Whenever Distinguishable objects into indistinguishable boxes $\rightarrow$Stirling number of second kind 

$\begin{Bmatrix} 4\\ 1 \end{Bmatrix}$ + $\begin{Bmatrix} 4\\ 2 \end{Bmatrix}$ + $\begin{Bmatrix} 4\\ 3 \end{Bmatrix}$  = 1 + 7 + 6 = 14

$\begin{Bmatrix} k\\ 1 \end{Bmatrix}$ = 1,

$\begin{Bmatrix} k\\ 2 \end{Bmatrix}$ = $2^{k-1} -1$

$\begin{Bmatrix} k\\ k-1 \end{Bmatrix}$ = $\binom{k}{2}$

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Use multinomial theorem for identical objects.

number of ways of distributing n identical objects into r persons= Combination of like objects

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we know distinguishable vs distinguishable - starling method is applicable

Is for distinguishable vs indistinguishable also - Starling method applicable?

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Starling method is used when there are different objects and identical boxes.

Answer 1

Here offices are identical 

We can put 4 employees into 1 office in 1 way  ⇒  [ $^4C_4$ = 1]

OR

We can put 3 employees into 1 office & 1 employee into another office in 4 ways ⇒  [ $^4C_3 $ = 4]

OR

We can put 2 employees into 1 office and 1 employee into another office and remaining 1 employee into another office in 6 ways ⇒  [ $^4C_2$ = 6]

OR 

We can put 2 employees into 1 office and 2 employees into another office in 3 ways ⇒  [ $\dfrac{^4C_2}{2}$ = $\dfrac{6}{2}$ = 3]

So, total no. of ways = 1 + 4 + 6 + 3 = 14 ways

Comments

should not the answer  be 3^(4)=81

each different person has 3 options as it can go to any of three offices

or can we look at this problem of partitioning of sum 4 in different ways

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Answer is 81 if offices are distinguishable

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we know distinguishable vs distinguishable - starling method is applicable

Is for distinguishable vs indistinguishable also - Starling method applicable?

Answer 2

14 will be the answer

Answer 3

This  question can be considered as division of 4 distinct items into three groups.

one way of grouping - {4},{0},{0} no of ways= 1

second way - {3},{1},{0} no of ways=4!/3!

third way - {2},{2},{0} no of ways=4!/2!2!*1/2!

fourth way={1},{1},{2} no of ways=4!/2!2!

Total ways=14