Question 1
In how many ways can 5 distinct items be distributed into 3 distinct boxes? (Empty boxes are allowed)
Step-by-Step Solution:
Concept: Distribution of distinct objects to distinct boxes - each object independently chooses a box.
Given:
- Distinct items: 5
- Distinct boxes: 3
- Empty boxes: Allowed
Strategy: Each item makes an independent choice of which box to go into.
Analysis:
- Item 1 can go into any of 3 boxes: 3 choices
- Item 2 can go into any of 3 boxes: 3 choices
- Item 3 can go into any of 3 boxes: 3 choices
- ...and so on for all 5 items
Formula: (number of boxes)^(number of items) = 3^5
Calculation:
Total ways = 3^5 = 243
Key Distinction:
This is NOT a permutation or combination problem! It's a direct application of the multiplication principle.
Related Scenarios:
1. No empty boxes allowed (Surjection):
Use Inclusion-Exclusion: 3! × S(5,3)
where S(n,k) is the Stirling number of second kind
2. Identical items, distinct boxes:
This becomes a "stars and bars" problem: C(5+3-1, 3-1)
3. Distinct items, identical boxes (Partition):
Use Stirling numbers: Σ S(5,k) for k=1 to 3
Verification:
- With 1 box: answer should be 1 (all items in one box)
- With 5 boxes, one item: answer should be 5
- Our answer 243 is reasonable: each item independently chooses from 3 options
Common Error: Don't use factorial or combination formulas here - items are making independent simultaneous choices.
Concept: Distribution of distinct objects to distinct boxes - each object independently chooses a box.
Given:
- Distinct items: 5
- Distinct boxes: 3
- Empty boxes: Allowed
Strategy: Each item makes an independent choice of which box to go into.
Analysis:
- Item 1 can go into any of 3 boxes: 3 choices
- Item 2 can go into any of 3 boxes: 3 choices
- Item 3 can go into any of 3 boxes: 3 choices
- ...and so on for all 5 items
Formula: (number of boxes)^(number of items) = 3^5
Calculation:
Total ways = 3^5 = 243
Key Distinction:
This is NOT a permutation or combination problem! It's a direct application of the multiplication principle.
Related Scenarios:
1. No empty boxes allowed (Surjection):
Use Inclusion-Exclusion: 3! × S(5,3)
where S(n,k) is the Stirling number of second kind
2. Identical items, distinct boxes:
This becomes a "stars and bars" problem: C(5+3-1, 3-1)
3. Distinct items, identical boxes (Partition):
Use Stirling numbers: Σ S(5,k) for k=1 to 3
Verification:
- With 1 box: answer should be 1 (all items in one box)
- With 5 boxes, one item: answer should be 5
- Our answer 243 is reasonable: each item independently chooses from 3 options
Common Error: Don't use factorial or combination formulas here - items are making independent simultaneous choices.