1. **Problem Statement:** Woodland College has two rows of 8 seats each (total 16 seats). There are 12 people to be seated. Among them, 3 want to sit in the first row and 6 refuse to sit in the first row. We need to find how many ways these 12 people can be seated under these conditions. 2. **Understanding the problem:** - Total seats: 16 (8 in first row, 8 in second row) - Total people: 12 - 3 people want to sit in the first row (must be in first row) - 6 people refuse to sit in the first row (must be in second row) - Remaining 3 people have no seating restrictions 3. **Step 1: Seat the 3 people who want the first row.** - They must occupy 3 of the 8 seats in the first row. - Number of ways to choose these 3 seats: $\binom{8}{3}$ - Number of ways to arrange these 3 people in those seats: $3!$ 4. **Step 2: Seat the 6 people who refuse the first row.** - They must occupy 6 of the 8 seats in the second row. - Number of ways to choose these 6 seats: $\binom{8}{6}$ - Number of ways to arrange these 6 people in those seats: $6!$ 5. **Step 3: Seat the remaining 3 people with no restrictions.** - Total remaining seats: $16 - 3 - 6 = 7$ seats (5 in first row + 2 in second row) - Number of ways to choose 3 seats out of these 7: $\binom{7}{3}$ - Number of ways to arrange these 3 people in those seats: $3!$ 6. **Step 4: Calculate total number of ways.** - Multiply all the above: $$\binom{8}{3} \times 3! \times \binom{8}{6} \times 6! \times \binom{7}{3} \times 3!$$ 7. **Step 5: Compute the values:** - $\binom{8}{3} = \frac{8!}{3!5!} = 56$ - $3! = 6$ - $\binom{8}{6} = \frac{8!}{6!2!} = 28$ - $6! = 720$ - $\binom{7}{3} = \frac{7!}{3!4!} = 35$ - $3! = 6$ 8. **Step 6: Multiply all:** $$56 \times 6 \times 28 \times 720 \times 35 \times 6$$ 9. **Step 7: Simplify stepwise:** - $56 \times 6 = 336$ - $336 \times 28 = 9408$ - $9408 \times 720 = 6,774,720$ - $6,774,720 \times 35 = 237,115,200$ - $237,115,200 \times 6 = 1,422,691,200$ **Final answer:** There are $1,422,691,200$ ways to seat the 12 people under the given conditions.