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How many ways are there to Distribute 7 distinct objects to 3 Distinct boxes and

  1. No box should be Empty
  2. Any box can be Empty
in Combinatory retagged by
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4 Comments

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Part 1:-

No box should be empty implies each box should have at least one object . This is same as number onto function $f:A->B$ 

where $|A|=7$ and $|B|=3$

Number of onto function = $3^{7}-3*2^{7}+3=1806$


Part 2:-

Number of arrangement where any box can be empty= total number of arrangement $-$ number of arrangement with no box is empty

Number of arrangement where any box can be empty= $3^{7}-1806=381$

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In part 2

Any of the box can be empty can be treated as No of functions from A → B right

so 3^7 should be the answer

its asked “ any box CAN be empty ” means there are no restrictions in the distribution right
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The wording “any box can be empty “ is ambiguous to me . As per me i have given solution based on at least one box is empty .

Your reasoning is also seems correct to me .
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this answer seems to be more convincing
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