| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve this problem, we can use the principle of inclusion-exclusion. First, we will calculate the total number of ways to choose a committee of 4 people from the group of 10 people without any restrictions, and then we will subtract the number of ways that violate the given conditions. 1. Calculate the total number of ways to choose a committee of 4 people from the group of 10 people without any restrictions: The number of ways to choose 4 people from 10 is given by the combination formula C(n, k) = n! / (k!(n-k)!), where n is the total number of people and k is the number of people to be chosen. In this case, n = 10 and k = 4. So, the total number of ways to choose a committee of 4 people from the group of 10 people without any restrictions is C(10, 4) = 10! / (4!(10-4)!) = 210. 2. Calculate the number of ways that violate the first condition (A and B should not be chosen together): If A and B are both chosen, then we have to choose 2 more people from the remaining 8 people. The number of ways to choose 2 people from 8 is given by the combination formula C(8, 2) = 8! / (2!(8-2)!) = 28. 3. Calculate the number of ways that violate the second condition (A, C, and F should not be chosen together): If A, C, and F are all chosen, then we have to choose 1 more person from the remaining 7 people. The number of ways to choose 1 person from 7 is given by the combination formula C(7, 1) = 7! / (1!(7-1)!) = 7. 4. Calculate the number of ways that violate both conditions: If A, B, C, and F are all chosen, then we have to choose 0 more people from the remaining 6 people. The number of ways to choose 0 people from 6 is given by the combination formula C(6, 0) = 6! / (0!(6-0)!) = 1. Using the principle of inclusion-exclusion, we can calculate the number of ways to choose a committee of 4 people that satisfies the given conditions: Total number of ways = Total number of ways without any restrictions - Number of ways that violate the first condition - Number of ways that violate the second condition + Number of ways that violate both conditions = 210 - 28 - 7 + 1 = 176. Therefore, the answer is 176. |
176 |
| deepseekr1 | To solve the problem of choosing a committee of 4 people from 10 individuals {A, B, C, D, E, F, G, H, I, J} with the given restrictions, we use the principle of inclusion-exclusion. 1. **Total number of ways without restrictions**: \[ \binom{10}{4} = 210 \] 2. **Subtract committees violating the first restriction (A and B together)**: \[ \binom{8}{2} = 28 \] 3. **Subtract committees violating the second restriction (A, C, F together)**: \[ \binom{7}{1} = 7 \] 4. **Add back the overlap (committees violating both restrictions)**: - The only committee violating both restrictions is {A, B, C, F}, which is 1 committee. Using inclusion-exclusion principle: \[ \text{Total invalid committees} = 28 + 7 - 1 = 34 \] 5. **Calculate valid committees**: \[ 210 - 34 = 176 \] Thus, the number of ways to choose the committee is \(\boxed{176}\). |
176 |
| gpto1 | To solve this problem, we need to find the number of ways to choose a committee of 4 people from a group of 10, under the given constraints: 1. **A and B should not be chosen together.** 2. **A, C, and F should not be chosen together.** We will break down the problem by considering whether person **A** is included in the committee or not. ### **Case 1:** A is **not included** in the committee. Since A is not on the committee, both constraints are automatically satisfied: - Constraint 1 is satisfied because A and B are not together. - Constraint 2 is satisfied because A is not present. We need to choose 4 people from the remaining 9 people (excluding A). Number of ways in Case 1: \[ \binom{9}{4} = \frac{9!}{4!(9 - 4)!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126 \] ### **Case 2:** A is **included** in the committee. Since A is on the committee, we need to choose 3 more people to complete the committee of 4. Constraints to consider: - **Constraint 1:** A and B cannot be together. Since A is included, B cannot be on the committee. - **Constraint 2:** A, C, and F cannot all be together. Since A is included, we must ensure that **C and F are not both included** with A. We are choosing 3 people from the remaining 8 people (excluding B, since B cannot be included): \[ \text{Available members: } \{C, D, E, F, G, H, I, J\} \] We need to avoid the selection where both C and F are included along with A. Total ways to choose 3 people from 8: \[ \binom{8}{3} = 56 \] Number of ways where both C and F are selected: - Fix C and F (both selected). - Choose 1 more person from the remaining 6 people: \[ \{D, E, G, H, I, J\} \] - Number of such ways: \[ \binom{6}{1} = 6 \] Therefore, the number of valid committees in Case 2: \[ 56 - 6 = 50 \] ### **Total number of valid committees:** \[ \text{Total ways} = \text{Ways in Case 1} + \text{Ways in Case 2} \] \[ \text{Total ways} = 126 + 50 = 176 \] **Answer:** 176 |
176 |
| GPT-4o | To solve this problem, we need to count the number of ways to choose a committee of 4 people from a group of 10 people such that (1) A and B are not chosen together, and (2) A, C, and F are not chosen together. First, let's calculate the total number of ways to choose a committee of 4 people from 10 people without any restrictions. This can be done using the binomial coefficient: \[ \binom{10}{4} = \frac{10!}{4!(10-4)!} = 210 \] Next, we need to count the number of ways that violate the given conditions and subtract those from the total. ### Condition 1: A and B should not be chosen together #### Case 1: A and B are both chosen: If A and B are both chosen, we need to choose 2 more people from the remaining 8 people (C, D, E, F, G, H, I, J): \[ \binom{8}{2} = \frac{8!}{2!(8-2)!} = 28 \] ### Condition 2: A, C, and F should not be chosen together #### Case 2: A, C, and F are all chosen: If A, C, and F are all chosen, we need to choose 1 more person from the remaining 7 people (B, D, E, G, H, I, J): \[ \binom{7}{1} = 7 \] ### Combined condition: A and B should not be chosen together and A, C, and F should not be chosen together #### Case 3: A, B, C, and F are all chosen: If A, B, C, and F are all chosen, this set violates both conditions. However, this is a subset of Case 1 and Case 2, so we must ensure we do not double-count this scenario. There is only 1 way for this to happen: \[ \binom{4}{4} = 1 \] Now, we use the principle of inclusion and exclusion to find the number of violating combinations: \[ \text{Violating combinations} = (\text{Case 1}) + (\text{Case 2}) - (\text{Case 3}) = 28 + 7 - 1 = 34 \] Finally, subtract the number of violating combinations from the total number of combinations: \[ \text{Valid combinations} = \binom{10}{4} - \text{Violating combinations} = 210 - 34 = 176 \] Therefore, the number of ways to choose a committee of 4 people such that A and B are not chosen together, and A, C, and F are not chosen together is: \[ \boxed{176} \] |
176 |
| llama3.1-70B-Instruct | To solve this problem, we will use the principle of inclusion-exclusion. First, let's find the total number of ways to choose a committee of 4 people from the group of 10 people without any restrictions. This is given by the combination formula: $$\binom{10}{4} = \frac{10!}{4!6!} = 210$$ Now, let's count the number of ways to choose a committee that violates the first restriction, i.e., A and B are chosen together. We can choose the remaining 2 people from the remaining 8 people in $\binom{8}{2} = 28$ ways. Next, let's count the number of ways to choose a committee that violates the second restriction, i.e., A, C, and F are chosen together. We can choose the remaining 1 person from the remaining 7 people in $\binom{7}{1} = 7$ ways. However, we have overcounted the number of committees that violate both restrictions. To correct this, we need to subtract the number of committees that violate both restrictions. This is given by the number of ways to choose the remaining 1 person from the remaining 7 people, which is $\binom{7}{1} = 7$. Therefore, the total number of ways to choose a committee that violates at least one of the restrictions is: $$28 + 7 - 7 = 28$$ Finally, we subtract this from the total number of ways to choose a committee without any restrictions to get the number of ways to choose a committee that satisfies both restrictions: $$210 - 28 = \boxed{182}$$ Therefore, there are 182 ways to choose a committee of 4 people from the group of 10 people that satisfies both restrictions. |
176 |
| llama3.1-405B-Instruct | To solve this problem, we can use the principle of inclusion-exclusion. First, let's find the total number of ways to choose a committee of 4 people from the group of 10 without any restrictions. This can be done using the combination formula: $$\binom{10}{4} = \frac{10!}{4!6!} = 210$$ Now, let's consider the restrictions: 1. A and B should not be chosen together. This means that we need to subtract the number of committees that include both A and B. There are $\binom{8}{2} = 28$ ways to choose the remaining 2 members of the committee from the remaining 8 people (excluding A and B). 2. A, C, and F should not be chosen together. This means that we need to subtract the number of committees that include all three of A, C, and F. There is only 1 way to choose the remaining member of the committee from the remaining 7 people (excluding A, C, and F). However, we have double-counted the committees that include both A and B, as well as A, C, and F. To correct for this, we need to add back the number of committees that include all four of A, B, C, and F. There are no such committees, since we are choosing a committee of 4 people and we have already excluded 4 people (A, B, C, and F). Therefore, the total number of ways to choose a committee of 4 people from the group of 10, subject to the given restrictions, is: $$210 - 28 - 1 + 0 = 181$$ So, there are 181 ways to choose such a committee. |
176 |
| DeepSeek-Coder-V2 | To solve this problem, we need to consider the constraints given and count the number of valid committees that can be formed under these constraints. Let's break down the problem step by step. ### Step 1: Total number of ways to choose a committee of 4 people from 10 First, we calculate the total number of ways to choose 4 people out of 10 without any constraints. This can be done using the combination formula: \[ \binom{10}{4} = \frac{10!}{4!(10-4)!} = 210 \] ### Step 2: Subtract the invalid combinations due to the constraints We have two constraints: 1. A and B should not be chosen together. 2. A, C, and F should not be chosen together. #### Constraint 1: A and B together If A and B are chosen together, we need to choose 2 more people from the remaining 8 people (C, D, E, F, G, H, I, J). The number of ways to do this is: \[ \binom{8}{2} = \frac{8!}{2!(8-2)!} = 28 \] #### Constraint 2: A, C, and F together If A, C, and F are chosen together, we need to choose 1 more person from the remaining 7 people (B, D, E, G, H, I, J). The number of ways to do this is: \[ \binom{7}{1} = \frac{7!}{1!(7-1)!} = 7 \] ### Step 3: Add back the over-counted combinations We need to consider the overlap between the two constraints. Specifically, if A, B, C, and F are all chosen together, this is an invalid combination that we have counted twice (once in each constraint). There are no valid ways to choose 4 people including A, B, C, and F together, so there is no need to add back any over-counted combinations. ### Step 4: Calculate the valid combinations Now, we subtract the invalid combinations from the total number of combinations: \[ \text{Valid combinations} = \binom{10}{4} - \binom{8}{2} - \binom{7}{1} = 210 - 28 - 7 = 175 \] Thus, the number of ways to choose a committee of 4 people from the group of 10 people, given the constraints, is: \[ \boxed{175} \] |
176 |
| Qwen2-72B-Instruct | To solve this problem, we can use the principle of inclusion-exclusion. First, we will calculate the total number of ways to choose a committee of 4 people from the group of 10 people without any restrictions, and then we will subtract the number of ways that violate the given conditions. 1. Calculate the total number of ways to choose a committee of 4 people from the group of 10 people without any restrictions: The number of ways to choose 4 people from 10 is given by the combination formula C(n, k) = n! / (k!(n-k)!), where n is the total number of people and k is the number of people to be chosen. In this case, n = 10 and k = 4. So, the total number of ways to choose a committee of 4 people from the group of 10 people without any restrictions is C(10, 4) = 10! / (4!(10-4)!) = 210. 2. Calculate the number of ways that violate the first condition (A and B should not be chosen together): If A and B are both chosen, then we have to choose 2 more people from the remaining 8 people. The number of ways to choose 2 people from 8 is given by the combination formula C(8, 2) = 8! / (2!(8-2)!) = 28. 3. Calculate the number of ways that violate the second condition (A, C, and F should not be chosen together): If A, C, and F are all chosen, then we have to choose 1 more person from the remaining 7 people. The number of ways to choose 1 person from 7 is given by the combination formula C(7, 1) = 7! / (1!(7-1)!) = 7. 4. Calculate the number of ways that violate both conditions: If A, B, C, and F are all chosen, then we have to choose 0 more people from the remaining 6 people. The number of ways to choose 0 people from 6 is given by the combination formula C(6, 0) = 6! / (0!(6-0)!) = 1. Using the principle of inclusion-exclusion, we can calculate the number of ways to choose a committee of 4 people that satisfies the given conditions: Total number of ways = Total number of ways without any restrictions - Number of ways that violate the first condition - Number of ways that violate the second condition + Number of ways that violate both conditions = 210 - 28 - 7 + 1 = 176. Therefore, the answer is 176. |
176 |
| baichuan4 | To solve this problem, we can use the principle of inclusion-exclusion. We need to find the total number of committees possible, then subtract the number of committees that violate the conditions (1) and (2). 1. Total number of committees of 4 people from a group of 10: This is simply the combination formula C(n, k), where n is the total number of people (10) and k is the size of the committee (4): C(10, 4) = 10! / (4! * (10 - 4)!) = 210 2. Number of committees that include both A and B: If A and B are already chosen, we only need to choose 2 more people from the remaining 8: C(8, 2) = 8! / (2! * (8 - 2)!) = 28 3. Number of committees that include A, C, and F: If A, C, and F are already chosen, we only need to choose 1 more person from the remaining 7: C(7, 1) = 7! / (1! * (7 - 1)!) = 7 4. Number of committees that include both A and B and also include A, C, and F: This is only possible if all 5 people (A, B, C, F and one more) are chosen. So, we need to choose 1 more person from the remaining 5: C(5, 1) = 5! / (1! * (5 - 1)!) = 5 Now, using the principle of inclusion-exclusion, we subtract the number of committees that violate condition (1) or (2) or both from the total number of committees: Total - (Committees with A and B) - (Committees with A, C, F) + (Committees with A, B, C, F) 210 - 28 - 7 + 5 = 170 So, there are 170 ways to choose a committee of 4 people satisfying the given conditions. |
176 |
| doubao | First find the total number of ways to choose a committee of 4 people from 10, which is 10C4 = (10*9*8*7)/(4*3*2*1) = 210. Now consider the cases where A and B are together. If A and B are in the committee, then we need to choose 2 more people from the remaining 8 (excluding A and B), which can be done in 8C2 = (8*7)/(2*1) = 28 ways. Next consider the case where A, C, and F are together. If A, C, and F are in the committee, then we need to choose 1 more person from the remaining 7 (excluding A, C, F), which can be done in 7C1 = 7 ways. But we have double-counted the case where both A and B are together and A, C, and F are also together. This case can be counted by choosing 1 more person from the remaining 5 (excluding A, B, C, F), which can be done in 5C1 = 5 ways. So the total number of invalid cases is 28 + 7 - 5 = 30. Finally, the number of valid ways to choose the committee is 210 - 30 = 180. |
176 |
| minimax | To solve this problem, we can use the principle of inclusion-exclusion. We will first calculate the total number of ways to choose a committee of 4 people from 10 without any restrictions, and then subtract the number of ways that violate the given conditions. Step 1: Calculate the total number of ways to choose a committee of 4 people from 10. This can be done using the combination formula, which is given by: \[ C(n, k) = \frac{n!}{k!(n-k)!} \] where \( n \) is the total number of items, \( k \) is the number of items to choose, and \( ! \) denotes factorial. So, the total number of ways to choose 4 people from 10 is: \[ C(10, 4) = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!} \] \[ C(10, 4) = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 \] Step 2: Calculate the number of ways that violate the first condition (A and B are chosen together). If A and B are chosen together, we then need to choose 2 more people from the remaining 8 people (excluding A and B). \[ C(8, 2) = \frac{8!}{2!(8-2)!} = \frac{8!}{2!6!} \] \[ C(8, 2) = \frac{8 \times 7}{2 \times 1} = 28 \] Step 3: Calculate the number of ways that violate the second condition (A, C, F are chosen together). If A, C, and F are chosen together, we then need to choose 1 more person from the remaining 7 people (excluding A, C, and F). \[ C(7, 1) = \frac{7!}{1!(7-1)!} = \frac{7!}{1!6!} \] \[ C(7, 1) = 7 \] Step 4: Calculate the number of ways that violate both conditions (A and B are chosen together, and A, C, F are chosen together). However, if A, C, and F are chosen together, A and B cannot be chosen together, so we don't have to subtract anything extra for this case. Step 5: Apply the principle of inclusion-exclusion. We subtract the number of ways that violate the first condition and the number of ways that violate the second condition from the total number of ways to choose a committee. \[ \text{Number of valid committees} = C(10, 4) - C(8, 2) - C(7, 1) \] \[ \text{Number of valid committees} = 210 - 28 - 7 \] \[ \text{Number of valid committees} = 210 - 35 \] \[ \text{Number of valid committees} = 175 \] Therefore, there are 175 ways to choose a committee of 4 people from the group of 10, given the restrictions. |
176 |
| kimi | To solve this problem, we can use the principle of inclusion-exclusion. We will first calculate the total number of ways to choose a committee of 4 people from 10 without any restrictions, and then subtract the number of ways that violate the given conditions. 1. Total ways to choose 4 people from 10 without any restrictions: This is a combination problem, which can be calculated using the binomial coefficient: \[ \binom{10}{4} = \frac{10!}{4!(10-4)!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 \] 2. Ways to choose a committee with A and B together: If A and B are both chosen, we need to choose 2 more people from the remaining 8 (since A and B are already chosen, and we have 10 - 2 = 8 people left). So, the number of ways to do this is: \[ \binom{8}{2} = \frac{8!}{2!(8-2)!} = \frac{8 \times 7}{2 \times 1} = 28 \] 3. Ways to choose a committee with A, C, and F together: If A, C, and F are all chosen, we only need to choose 1 more person from the remaining 7 (since A, C, and F are already chosen, and we have 10 - 3 = 7 people left). So, the number of ways to do this is: \[ \binom{7}{1} = 7 \] However, we have to be careful not to double-count the scenario where both A and B are chosen together with A, C, and F. Since A, C, and F are together, B cannot be chosen with them, so there is no overlap between the two scenarios. 4. Using the principle of inclusion-exclusion: We subtract the number of ways that violate the first condition (A and B together) and the number of ways that violate the second condition (A, C, and F together) from the total number of ways: \[ \text{Total valid ways} = \binom{10}{4} - \binom{8}{2} - \binom{7}{1} \] \[ \text{Total valid ways} = 210 - 28 - 7 \] \[ \text{Total valid ways} = 175 \] Therefore, there are 175 ways to choose a committee of 4 people from the group of 10 with the given restrictions. |
176 |
| wenxinyiyan | 首先，从10人中选出4人组成委员会的总方案数为C(10,4)=210种。 然后，我们需要考虑需要排除的情况： * A和B同时被选的情况：如果A和B同时被选，那么剩下两个人可以从其余8人中选出，即C(8,2)=28种。 * A、C、F同时被选的情况：如果A、C、F同时被选，那么剩下一个人可以从其余7人中选出，即C(7,1)=7种。 注意，上述两种情况有重叠部分，即A、B、C、F四人同时被选的情况。这种情况只有1种。 因此，符合条件的选法应该是总选法减去这两种不符合条件的选法，同时加上重叠部分，即210-28-7+1=176种。 所以，总共有176种不同的选法来组成这样一个委员会。 |
176 |
| zhipu | To solve this problem, we will use the principle of inclusion-exclusion. We will first calculate the total number of ways to choose a committee of 4 people from the 10, and then we will subtract the number of ways that violate the given conditions. **Step 1: Calculate the total number of ways to choose a committee of 4 people from 10.** The number of ways to choose 4 people out of 10 is given by the combination formula \( \binom{n}{k} \), where \( n \) is the total number of items to choose from, and \( k \) is the number of items to choose. \[ \binom{10}{4} = \frac{10!}{4!(10-4)!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 \] **Step 2: Subtract the number of ways that violate the conditions.** **Condition 1: A and B should not be chosen together.** First, calculate the number of ways to choose a committee that includes both A and B. If A and B are both in the committee, we need to choose 2 more people from the remaining 8 people. \[ \binom{8}{2} = \frac{8!}{2!(8-2)!} = \frac{8 \times 7}{2 \times 1} = 28 \] **Condition 2: A, C, and F should not be chosen together.** Next, calculate the number of ways to choose a committee that includes A, C, and F. If A, C, and F are all in the committee, we need to choose 1 more person from the remaining 7 people. \[ \binom{7}{1} = \frac{7!}{1!(7-1)!} = 7 \] **Step 3: Apply the principle of inclusion-exclusion.** We need to ensure that we do not double-count the cases where both conditions are violated (i.e., a committee that includes A, B, C, and F). However, in this problem, it is impossible to violate both conditions simultaneously because if A, C, and F are together, B cannot be in the committee due to the size constraint (only 4 members allowed). Therefore, the number of ways to choose a committee that violates either condition is simply the sum of the individual violations: \[ 28 \text{ (A and B together)} + 7 \text{ (A, C, and F together)} = 35 \] **Step 4: Subtract the invalid combinations from the total.** \[ 210 \text{ (total ways)} - 35 \text{ (invalid ways)} = 175 \] **Conclusion:** There are 175 ways to choose a committee of 4 people from the group of 10 that satisfies the given conditions. \[ \boxed{175} \] |
176 |
| gpto1 | To solve this problem, we need to find the number of ways to choose a committee of 4 people from a group of 10, under the given constraints: 1. **A and B should not be chosen together.** 2. **A, C, and F should not be chosen together.** We will break down the problem by considering whether person **A** is included in the committee or not. ### **Case 1:** A is **not included** in the committee. Since A is not on the committee, both constraints are automatically satisfied: - Constraint 1 is satisfied because A and B are not together. - Constraint 2 is satisfied because A is not present. We need to choose 4 people from the remaining 9 people (excluding A). Number of ways in Case 1: \[ \binom{9}{4} = \frac{9!}{4!(9 - 4)!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126 \] ### **Case 2:** A is **included** in the committee. Since A is on the committee, we need to choose 3 more people to complete the committee of 4. Constraints to consider: - **Constraint 1:** A and B cannot be together. Since A is included, B cannot be on the committee. - **Constraint 2:** A, C, and F cannot all be together. Since A is included, we must ensure that **C and F are not both included** with A. We are choosing 3 people from the remaining 8 people (excluding B, since B cannot be included): \[ \text{Available members: } \{C, D, E, F, G, H, I, J\} \] We need to avoid the selection where both C and F are included along with A. Total ways to choose 3 people from 8: \[ \binom{8}{3} = 56 \] Number of ways where both C and F are selected: - Fix C and F (both selected). - Choose 1 more person from the remaining 6 people: \[ \{D, E, G, H, I, J\} \] - Number of such ways: \[ \binom{6}{1} = 6 \] Therefore, the number of valid committees in Case 2: \[ 56 - 6 = 50 \] ### **Total number of valid committees:** \[ \text{Total ways} = \text{Ways in Case 1} + \text{Ways in Case 2} \] \[ \text{Total ways} = 126 + 50 = 176 \] **Answer:** 176 |
176 |
| deepseekv3 | To solve this problem, we need to determine the number of ways to choose a committee of 4 people from a group of 10, subject to two constraints: 1. **A and B should not be chosen together.** 2. **A, C, and F should not be chosen together.** We will approach this problem using the principle of **inclusion-exclusion**. ### Step 1: Calculate the Total Number of Committees Without Any Constraints The total number of ways to choose a committee of 4 people from 10 is given by the combination formula: \[ \text{Total} = \binom{10}{4} = \frac{10!}{4! \cdot 6!} = 210 \] ### Step 2: Subtract the Number of Committees That Violate Constraint 1 (A and B are both chosen) First, count the number of committees where both A and B are chosen. If A and B are both on the committee, we need to choose 2 more members from the remaining 8 people: \[ \text{Violations of Constraint 1} = \binom{8}{2} = \frac{8!}{2! \cdot 6!} = 28 \] ### Step 3: Subtract the Number of Committees That Violate Constraint 2 (A, C, and F are all chosen) Next, count the number of committees where A, C, and F are all chosen. If A, C, and F are on the committee, we need to choose 1 more member from the remaining 7 people: \[ \text{Violations of Constraint 2} = \binom{7}{1} = 7 \] ### Step 4: Add Back the Number of Committees That Violate Both Constraints (A, B, C, and F are all chosen) Some committees may violate both constraints simultaneously (i.e., A, B, C, and F are all chosen). Such committees have been subtracted twice (once for each constraint), so we need to add them back once. If A, B, C, and F are all on the committee, no additional members are needed. There is only 1 such committee: \[ \text{Violations of Both Constraints} = 1 \] ### Step 5: Apply the Inclusion-Exclusion Principle Using the inclusion-exclusion principle, the number of valid committees is: \[ \text{Valid Committees} = \text{Total} - \text{Violations of Constraint 1} - \text{Violations of Constraint 2} + \text{Violations of Both Constraints} \] Substituting the values: \[ \text{Valid Committees} = 210 - 28 - 7 + 1 = 176 \] ### Final Answer The number of ways to choose a committee of 4 people from the group of 10, subject to the given constraints, is **176**. |
176 |