Question 1
In how many ways can 6 distinct items be distributed into 4 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: 6
- Distinct boxes: 4
- Empty boxes: Allowed
Strategy: Each item makes an independent choice of which box to go into.
Analysis:
- Item 1 can go into any of 4 boxes: 4 choices
- Item 2 can go into any of 4 boxes: 4 choices
- Item 3 can go into any of 4 boxes: 4 choices
- ...and so on for all 6 items
Formula: (number of boxes)^(number of items) = 4^6
Calculation:
Total ways = 4^6 = 4096
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: 4! × S(6,4)
where S(n,k) is the Stirling number of second kind
2. Identical items, distinct boxes:
This becomes a "stars and bars" problem: C(6+4-1, 4-1)
3. Distinct items, identical boxes (Partition):
Use Stirling numbers: Σ S(6,k) for k=1 to 4
Verification:
- With 1 box: answer should be 1 (all items in one box)
- With 6 boxes, one item: answer should be 6
- Our answer 4096 is reasonable: each item independently chooses from 4 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: 6
- Distinct boxes: 4
- Empty boxes: Allowed
Strategy: Each item makes an independent choice of which box to go into.
Analysis:
- Item 1 can go into any of 4 boxes: 4 choices
- Item 2 can go into any of 4 boxes: 4 choices
- Item 3 can go into any of 4 boxes: 4 choices
- ...and so on for all 6 items
Formula: (number of boxes)^(number of items) = 4^6
Calculation:
Total ways = 4^6 = 4096
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: 4! × S(6,4)
where S(n,k) is the Stirling number of second kind
2. Identical items, distinct boxes:
This becomes a "stars and bars" problem: C(6+4-1, 4-1)
3. Distinct items, identical boxes (Partition):
Use Stirling numbers: Σ S(6,k) for k=1 to 4
Verification:
- With 1 box: answer should be 1 (all items in one box)
- With 6 boxes, one item: answer should be 6
- Our answer 4096 is reasonable: each item independently chooses from 4 options
Common Error: Don't use factorial or combination formulas here - items are making independent simultaneous choices.