If n is the number of ways five different employees can sit into four indistinguishable offices where any office may have any number of persons including zero, then n is equal to:
- 47
- 53
- 51
- 43
The Correct Option is C
Approach Solution - 1
Total ways to partition 5 into 4 parts are:
- 5, 0, 0, 0 → 1 way
- 4, 1, 0, 0 → \(\frac{5!}{4!} = 5 \text{ ways}\)
- 3, 2, 0, 0 →\(\frac{5!}{3!2!} = 10 \text{ ways}\)
- 2, 2, 1, 0 →\(\frac{5!}{2!2!1!} = 15 \text{ ways}\)
- 2, 1, 1, 1 →\(\frac{5!}{2!1!1!1!} = 10 \text{ ways}\)
- 3, 1, 1, 0 → \(\frac{5!}{3!1!1!} = 10 \text{ ways}\)
Total:
\[ 1 + 5 + 10 + 15 + 10 + 10 = 51 \text{ ways} \]
Approach Solution -2
Find the number of ways to distribute 5 distinct employees into 4 indistinguishable offices, with empty offices allowed.
Concept Used:
When offices are indistinguishable, we are essentially partitioning the set of 5 distinct employees into at most 4 unlabeled subsets (since offices are indistinguishable). This is equivalent to counting the number of ways to partition a set of 5 distinct objects into at most 4 non-empty unlabeled subsets, which is given by the Bell number \(B_5\) minus the number of partitions into 5 subsets (which would require 5 offices, but we only have 4). However, since empty offices are allowed, we can use between 1 and 4 non-empty subsets.
Step-by-Step Solution:
Step 1: Interpret the problem.
We have 5 distinct employees and 4 indistinguishable offices. Any office can have any number of persons, including zero. This means we are partitioning the set of 5 employees into at most 4 unlabeled subsets (some offices may be empty).
Step 2: Count the number of partitions.
Let \(S(5,k)\) denote the Stirling numbers of the second kind (number of ways to partition 5 distinct objects into k non-empty unlabeled subsets).
We need number of partitions into 1, 2, 3, or 4 non-empty subsets (since we have 4 offices, we can use 1, 2, 3, or 4 of them).
Step 3: Recall Stirling numbers of the second kind for n=5:
\[ S(5,1) = 1 \quad (\text{all in one office}) \] \[ S(5,2) = 15 \] \[ S(5,3) = 25 \] \[ S(5,4) = 10 \] \[ S(5,5) = 1 \]Step 4: Compute total number of distributions.
Since offices are indistinguishable, the number of ways is:
\[ n = S(5,1) + S(5,2) + S(5,3) + S(5,4) \] \[ n = 1 + 15 + 25 + 10 = 51 \]Step 5: Verify alternative approach.
Total number of partitions of 5 elements = Bell number \(B_5 = S(5,1)+S(5,2)+S(5,3)+S(5,4)+S(5,5) = 1+15+25+10+1 = 52\).
We exclude the case where all 5 are in separate offices (5 non-empty subsets) because that would require 5 offices, but we only have 4. So \(n = B_5 - S(5,5) = 52 - 1 = 51\).
Therefore, the number of ways \(n\) is 51.
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