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In how many ways can 4 balls be placed in 4 boxes if the balls are distinguishable, and the boxes are indistinguishable?

 Aug 27, 2022
 #1
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This is calculated using "Stirling Numbers of the Second Kind"

 

k=4; n=4; (1/k! * sumfor(i, 0, k,(-1)^i * (k nCr  i) * (k - i)^n) * k!

 

==24 different ways - without any restrictions

 Aug 27, 2022
edited by Guest  Aug 27, 2022
 #2
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Hey I don't think 24 would be the correct answer
Also do you think there is another way to do it? I've never seen this before.

 Aug 27, 2022
 #3
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Go to this link and see the formula used (Formula(9)) under "distinct"   and   "identical" :

 

https://www.careerbless.com/aptitude/qa/permutations_combinations_imp7.php

Guest Aug 27, 2022
 #4
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There are 5 cases: 

 

4 - 0 - 0 - 0 

3 - 1 - 0 - 0 

2 - 2 - 0 - 0

2 - 1 - 1 - 0

1 - 1 - 1 - 1

 

For the first case, there is only 1 way. 

 

In the second case, there are \({4 \choose 3} = 4 \) ways.

 

In the third case, there are \({4 \choose 2} = 6\) ways.  

 

In the fourth case, there are \({4 \choose 2} \times 2 = 12\) ways. 

 

For the fifth case, there is only 1 way.

 

So, there are \(1 + 4 + 6 + 12 + 1 = \color{brown}\boxed{24}\) ways. 

 Aug 28, 2022

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