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

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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

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