1. **Stating the problem:** We have 6 Democrats and 4 Republicans from the same city in the House of Representatives. We want to find the probability that all four leadership positions are filled by representatives from this city. 2. **Understanding the problem:** The total number of representatives from the city is $6 + 4 = 10$. We want to select 4 leaders all from these 10 representatives. 3. **Formula used:** The probability that all four leadership positions are from the city is the number of ways to choose 4 leaders from the 10 city representatives divided by the total number of ways to choose 4 leaders from the entire House. Let $N$ be the total number of representatives in the House (not given explicitly, so we assume the problem focuses only on the city representatives). If the total House size is unknown, the problem implies selecting all 4 leaders from the city representatives out of the total 4 leadership positions. 4. **Calculating the probability:** Since all 4 leadership positions must be from the city representatives, and there are exactly 10 city representatives, the number of ways to choose 4 leaders from these 10 is: $$\binom{10}{4}$$ If the total number of representatives in the House is $T$, the total number of ways to choose 4 leaders is: $$\binom{T}{4}$$ 5. **Final probability:** $$P = \frac{\binom{10}{4}}{\binom{T}{4}}$$ 6. **Interpretation:** Without the total number $T$ of representatives in the House, we cannot compute a numeric probability. If the problem assumes the House consists only of these 10 representatives, then: $$P = \frac{\binom{10}{4}}{\binom{10}{4}} = 1$$ Otherwise, the probability depends on $T$. **Summary:** The probability that all four leadership positions are from the city representatives is: $$P = \frac{\binom{10}{4}}{\binom{T}{4}}$$ where $T$ is the total number of representatives in the House.