Answer: sum from k=0 to n of (-1)k × C(n,k) × (n-k)m.
- A nm, the count of all functions from A to B without restriction
- B sum from k=0 to n of (-1)k × C(n,k) × (n-k)m
- C n! × C(m,n), mistakenly mixing a permutation count with a combination
- D mn, swapping the roles of the domain and codomain sizes
Correct answer: B. sum from k=0 to n of (-1)k × C(n,k) × (n-k)m
Explanation: Inclusion-exclusion over elements of B left uncovered: Σₖ₌₀ⁿ (−1)ᵏ C(n,k)(n−k)ᵐ. Each term accounts for functions missing exactly k elements of B.
Counting tree for selecting 2 items from {A, B, C} without repetition: 3 x 2 = 6 ordered arrangements.
Concept context
Counting, arrangements, and selections