1. **Problem statement:** We need to form a group of 5 people from 3 married couples (6 people), 4 male staff, and 5 female staff, with the condition that no married couple is included together in the group. 2. **Understanding the problem:** The total people are 3 couples (6 people), 4 male staff, and 5 female staff, so total $6 + 4 + 5 = 15$ people. 3. **Key rule:** No couple can be selected together. This means if one member of a couple is selected, the other cannot be. 4. **Approach:** - From each couple, at most one person can be selected. - The 4 male and 5 female staff are independent and can be selected freely. 5. **Step 1: Count the number of ways to select people from the couples with no couple together.** - Each couple has 2 people, so from 3 couples, we can select 0, 1, 2, or 3 people but not both from the same couple. - Number of ways to select $k$ people from 3 couples with no couple together is $\binom{3}{k} \times 2^k$ because we choose which $k$ couples to pick from and for each chosen couple, select 1 of 2 people. 6. **Step 2: Let $k$ be the number of people selected from couples. Then the remaining $5-k$ people must be selected from the 9 staff (4 male + 5 female).** - Number of ways to select $5-k$ people from 9 staff is $\binom{9}{5-k}$. 7. **Step 3: Total number of ways is sum over $k=0$ to $3$ of:** $$\binom{3}{k} \times 2^k \times \binom{9}{5-k}$$ 8. **Calculate each term:** - For $k=0$: $\binom{3}{0} \times 2^0 \times \binom{9}{5} = 1 \times 1 \times 126 = 126$ - For $k=1$: $\binom{3}{1} \times 2^1 \times \binom{9}{4} = 3 \times 2 \times 126 = 756$ - For $k=2$: $\binom{3}{2} \times 2^2 \times \binom{9}{3} = 3 \times 4 \times 84 = 1008$ - For $k=3$: $\binom{3}{3} \times 2^3 \times \binom{9}{2} = 1 \times 8 \times 36 = 288$ 9. **Step 4: Sum all:** $$126 + 756 + 1008 + 288 = 2178$$ 10. **Final answer:** The number of ways to form the group with no couple together is **2178**.