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. **Understanding the problem:** We have 6 distinct people and 2 distinct rows. Each person must be assigned to exactly one row, and all 6 people must be placed in one of the two rows. 3. **Formula and approach:** This is a problem of distributing distinct objects (people) into distinct groups (rows) with no one left out. - First, decide how many people go into the first row (say $k$), then the rest $(6-k)$ go into the second row. - For each $k$, choose which $k$ people go to the first row: $\binom{6}{k}$ ways. - Then arrange the $k$ people in the first row: $k!$ ways. - Arrange the remaining $(6-k)$ people in the second row: $(6-k)!$ ways. 4. **Calculate total ways:** Sum over all possible $k$ from 0 to 6: $$\sum_{k=0}^6 \binom{6}{k} k! (6-k)!$$ 5. **Simplify:** Note that $\binom{6}{k} k! (6-k)! = \frac{6!}{k!(6-k)!} \times k! \times (6-k)! = 6! = 720$ Since this is constant for each $k$, and $k$ runs from 0 to 6 (7 values), total ways = $7 \times 720 = 5040$. 6. **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:** $$\boxed{5040}$$