1. **Problem Statement:** A photographer wants to arrange 6 distinct people into 2 distinct rows for a photo. How many ways can this be done if all 6 people must be in one of the two rows? 2. **Formula and Explanation:** We need to arrange 6 distinct people into 2 distinct rows, with all people placed in one of the two rows. This means each person is assigned to exactly one row, and the order within each row matters (since it's a photo arrangement). The number of ways to distribute 6 distinct people into 2 distinct rows is: - First, decide how many people go into the first row (say $k$), where $k$ can be from 0 to 6. - For each $k$, choose which $k$ people go into the first row: $\binom{6}{k}$ ways. - Arrange the $k$ people in the first row: $k!$ ways. - Arrange the remaining $6-k$ people in the second row: $(6-k)!$ ways. So total ways for each $k$ is: $$\binom{6}{k} \times k! \times (6-k)!$$ Summing over all $k$ from 0 to 6: $$\sum_{k=0}^6 \binom{6}{k} k! (6-k)!$$ 3. **Simplification:** Note that $\binom{6}{k} = \frac{6!}{k!(6-k)!}$, so: $$\binom{6}{k} k! (6-k)! = \frac{6!}{k!(6-k)!} \times k! \times (6-k)! = 6!$$ Since this is true for each $k$, and $k$ runs from 0 to 6 (7 values), total ways: $$7 \times 6! = 7 \times 720 = 5040$$ 4. **Interpretation:** This means there are 5040 ways to arrange 6 distinct people into 2 distinct rows, considering all possible distributions of people between the rows. **Final answer:** $$\boxed{5040}$$