1. **Problem statement:** There are 40 members in a club including Ranuf and Saed. 35 members will travel in a coach and 5 in a car. Ranuf must be in the coach and Saed must be in the car. We need to find the number of ways to choose the members who will travel in the coach. 2. **Understanding the problem:** Since Ranuf is already fixed in the coach, we only need to choose the remaining members for the coach from the other 38 members (excluding Ranuf and Saed). 3. **Formula used:** The number of ways to choose $k$ members from $n$ members is given by the combination formula: $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$ 4. **Applying the formula:** - Total members excluding Ranuf and Saed: $40 - 2 = 38$ - Number of members to choose for the coach excluding Ranuf: $35 - 1 = 34$ So, the number of ways to choose the remaining coach members is: $$\binom{38}{34}$$ 5. **Simplification:** $$\binom{38}{34} = \binom{38}{4}$$ (since $\binom{n}{k} = \binom{n}{n-k}$) 6. **Calculating $\binom{38}{4}$:** $$\binom{38}{4} = \frac{38 \times 37 \times 36 \times 35}{4 \times 3 \times 2 \times 1}$$ 7. **Evaluating:** $$= \frac{38 \times 37 \times 36 \times 35}{24}$$ Calculate numerator: $$38 \times 37 = 1406$$ $$1406 \times 36 = 50616$$ $$50616 \times 35 = 1771560$$ Divide by denominator: $$\frac{1771560}{24} = 73815$$ **Final answer:** There are $73815$ ways to choose the members who will travel in the coach.