| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve this problem, we can use the multinomial coefficient formula. The multinomial coefficient is used when we have a set of n items and we want to divide them into k groups of sizes n1, n2, ..., nk, where n1 + n2 + ... + nk = n. In this case, we have a set of 10 people and we want to divide them into 3 groups of sizes 3, 4, and 3, respectively. We can use the multinomial coefficient formula: n! / (n1! * n2! * ... * nk!) where n is the total number of items, and n1, n2, ..., nk are the sizes of the groups. In this case, we have: n = 10 n1 = 3 n2 = 4 n3 = 3 So we can plug these values into the formula: 10! / (3! * 4! * 3!) Calculating the factorials: 10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 3,628,800 3! = 3 × 2 × 1 = 6 4! = 4 × 3 × 2 × 1 = 24 3! = 3 × 2 × 1 = 6 Now we can calculate the multinomial coefficient: 10! / (3! * 4! * 3!) = 3,628,800 / (6 * 24 * 6) = 3,628,800 / 864 = 4,200 So there are 4,200 ways to split the group of 10 people into 3 different committees of 3, 4, and 3 people, respectively. |
4200 |
| deepseekr1 | To determine the number of ways to split a group of 10 people into three different committees of 3, 4, and 3 people respectively, we need to consider whether the committees are labeled or unlabeled. The problem states that the committees are "different" and specifies their sizes "respectively," implying that they are labeled (i.e., each committee is distinct even if they have the same size). 1. **Choosing the first committee (3 people):** \[ \binom{10}{3} = 120 \] 2. **Choosing the second committee (4 people) from the remaining 7:** \[ \binom{7}{4} = 35 \] 3. **Choosing the third committee (3 people) from the remaining 3:** \[ \binom{3}{3} = 1 \] Multiplying these values together gives the total number of ways to form the committees: \[ 120 \times 35 \times 1 = 4200 \] This calculation is equivalent to the multinomial coefficient for dividing 10 people into groups of sizes 3, 4, and 3: \[ \frac{10!}{3! \cdot 4! \cdot 3!} = 4200 \] Since the committees are considered labeled (different), we do not need to divide by any factorial to account for indistinctness. Therefore, the number of ways to split the group is: \[ \boxed{4200} \] |
4200 |
| gpto1 | To solve this problem, we need to determine the number of ways to divide a group of 10 people into three distinct committees of sizes 3, 4, and 3, respectively. Since the committees are different (they are distinguishable), we use the concept of **multinomial coefficients** from combinatorics. ### Understanding the Problem: - **Total people**: 10 - **Committee sizes**: 3, 4, and 3 - **Committees are distinguishable**: This means that committee A (size 3) is different from committee B (size 4), which is different from committee C (size 3). The order and labeling of committees matter. ### Multinomial Coefficient Formula: The number of ways to partition a set of \( n \) distinct elements into \( r \) distinguishable groups of sizes \( n_1, n_2, \dots, n_r \) (where \( n_1 + n_2 + \dots + n_r = n \)) is given by the **multinomial coefficient**: \[ \text{Number of ways} = \dfrac{n!}{n_1! \times n_2! \times \dots \times n_r!} \] ### Applying the Formula: Let's assign: - \( n = 10 \) (total people) - \( n_1 = 3 \) (size of the first committee) - \( n_2 = 4 \) (size of the second committee) - \( n_3 = 3 \) (size of the third committee) Plugging these values into the multinomial coefficient formula: \[ \text{Number of ways} = \dfrac{10!}{3! \times 4! \times 3!} \] ### Calculating the Factorials: First, compute the factorials needed: - \( 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 3,628,800 \) - \( 3! = 3 \times 2 \times 1 = 6 \) - \( 4! = 4 \times 3 \times 2 \times 1 = 24 \) Now compute the denominator: \[ \text{Denominator} = 3! \times 4! \times 3! = 6 \times 24 \times 6 = 864 \] ### Performing the Division: Now divide the numerator by the denominator: \[ \text{Number of ways} = \dfrac{3,628,800}{864} \] To simplify the calculation, we can perform the division step by step. #### Simplifying the Division: 1. **Divide numerator and denominator by 16** (since 16 is a common factor of both 3,628,800 and 864): \[ \dfrac{3,628,800 \div 16}{864 \div 16} = \dfrac{226,800}{54} \] 2. **Divide numerator and denominator by 18** (since 18 divides both 226,800 and 54): \[ \dfrac{226,800 \div 18}{54 \div 18} = \dfrac{12,600}{3} \] 3. **Divide numerator and denominator**: \[ \dfrac{12,600}{3} = 4,200 \] Alternatively, you can perform the division directly: \[ \text{Number of ways} = \dfrac{3,628,800}{864} = 4,200 \] ### Conclusion: Therefore, there are **4,200 ways** to split the group of 10 people into three distinguishable committees of sizes 3, 4, and 3. **Answer:** **4,200** |
4200 |
| GPT-4o | To determine the number of ways to split a group of 10 people into three different committees of 3, 4, and 3 people, we need to approach this problem using combinatorial methods. 1. **Choose 3 people out of 10 for the first committee:** The number of ways to choose 3 people from 10 is given by the combination formula \(\binom{10}{3}\): \[ \binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10!}{3! \cdot 7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \] 2. **Choose 4 people out of the remaining 7 for the second committee:** The number of ways to choose 4 people from the remaining 7 is given by \(\binom{7}{4}\): \[ \binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7!}{4! \cdot 3!} = \frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1} = 35 \] 3. **Choose 3 people out of the remaining 3 for the third committee:** The number of ways to choose 3 people from the remaining 3 is given by \(\binom{3}{3}\): \[ \binom{3}{3} = \frac{3!}{3!(3-3)!} = \frac{3!}{3! \cdot 0!} = 1 \] Now, we need to multiply these choices together to find the total number of ways to form the committees: \[ 120 \times 35 \times 1 = 4200 \] However, this calculation considers the committees as distinguishable. If the committees are considered indistinguishable, we have overcounted the number of ways to form the committees. Since there are 3 committees and they are indistinguishable, we must divide by the number of permutations of these 3 committees, which is \(3!\): \[ 3! = 6 \] Therefore, the number of ways to split the group of 10 people into three committees of 3, 4, and 3 people, where the committees are indistinguishable, is: \[ \frac{4200}{6} = 700 \] So, the final answer is: \[ \boxed{700} \] |
4200 |
| llama3.1-70B-Instruct | To solve this problem, we can use the concept of combinations. We need to choose 3 people out of 10 for the first committee, then 4 people out of the remaining 7 for the second committee, and finally 3 people out of the remaining 3 for the third committee. The number of ways to choose 3 people out of 10 is given by the combination formula: C(10, 3) = 10! / (3! \* (10-3)!) = 10! / (3! \* 7!) = 120 The number of ways to choose 4 people out of the remaining 7 is: C(7, 4) = 7! / (4! \* (7-4)!) = 7! / (4! \* 3!) = 35 The number of ways to choose 3 people out of the remaining 3 is: C(3, 3) = 3! / (3! \* (3-3)!) = 1 Since these choices are independent, we can multiply the number of ways to get the total number of ways to form the committees: 120 \* 35 \* 1 = 4200 Therefore, there are 4200 ways to split the group of 10 people into 3 different committees of 3, 4, and 3 people, respectively. |
4200 |
| llama3.1-405B-Instruct | To solve this problem, we can use the concept of combinations. We need to choose 3 people out of 10 for the first committee, 4 people out of the remaining 7 for the second committee, and 3 people out of the remaining 3 for the third committee. The number of ways to choose 3 people out of 10 is given by the combination formula: C(10, 3) = 10! / (3! \* (10-3)!) = 10! / (3! \* 7!) = 120 The number of ways to choose 4 people out of the remaining 7 is: C(7, 4) = 7! / (4! \* (7-4)!) = 7! / (4! \* 3!) = 35 The number of ways to choose 3 people out of the remaining 3 is: C(3, 3) = 3! / (3! \* (3-3)!) = 3! / (3! \* 0!) = 1 Since the order of the committees does not matter, we need to divide the product of these combinations by the number of ways to arrange the committees, which is 3! = 6. Therefore, the total number of ways to split the group of 10 people into 3 committees of 3, 4, and 3 people is: (120 \* 35 \* 1) / 6 = 700 So, there are 700 ways to split the group of 10 people into 3 committees of 3, 4, and 3 people. |
4200 |
| DeepSeek-Coder-V2 | To solve the problem of splitting a group of 10 people into 3 different committees of 3, 4, and 3 people, respectively, we can use the concept of combinations and permutations. Here is the step-by-step solution: 1. **Choose 3 people for the first committee:** The number of ways to choose 3 people out of 10 is given by the combination formula \( \binom{10}{3} \): \[ \binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10!}{3! \cdot 7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \] 2. **Choose 4 people for the second committee from the remaining 7 people:** The number of ways to choose 4 people out of the remaining 7 is given by \( \binom{7}{4} \): \[ \binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7!}{4! \cdot 3!} = \frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1} = 35 \] 3. **The remaining 3 people will automatically form the third committee:** Since the remaining 3 people are already chosen, there is only 1 way to form this committee. 4. **Calculate the total number of ways to form the committees:** Since the choices are independent, we multiply the number of ways to form each committee: \[ 120 \times 35 \times 1 = 4200 \] However, we need to account for the fact that the order in which we form the committees does not matter. Since we have two committees of the same size (3 people each), we need to divide by the number of ways to arrange these two committees, which is 2! (since there are 2 ways to arrange 2 items): \[ \frac{4200}{2!} = \frac{4200}{2} = 2100 \] Therefore, the number of ways to split the group of 10 people into 3 different committees of 3, 4, and 3 people, respectively, is: \[ \boxed{2100} \] |
4200 |
| Qwen2-72B-Instruct | To solve this problem, we can use the multinomial coefficient formula. The multinomial coefficient is used when we have a set of n items and we want to divide them into k groups of sizes n1, n2, ..., nk, where n1 + n2 + ... + nk = n. In this case, we have a set of 10 people and we want to divide them into 3 groups of sizes 3, 4, and 3, respectively. We can use the multinomial coefficient formula: n! / (n1! * n2! * ... * nk!) where n is the total number of items, and n1, n2, ..., nk are the sizes of the groups. In this case, we have: n = 10 n1 = 3 n2 = 4 n3 = 3 So we can plug these values into the formula: 10! / (3! * 4! * 3!) Calculating the factorials: 10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 3,628,800 3! = 3 × 2 × 1 = 6 4! = 4 × 3 × 2 × 1 = 24 3! = 3 × 2 × 1 = 6 Now we can calculate the multinomial coefficient: 10! / (3! * 4! * 3!) = 3,628,800 / (6 * 24 * 6) = 3,628,800 / 864 = 4,200 So there are 4,200 ways to split the group of 10 people into 3 different committees of 3, 4, and 3 people, respectively. |
4200 |
| baichuan4 | To solve this problem, we can use the combination formula, which is given by: $$C(n, k) = \frac{n!}{k!(n-k)!}$$ where $n$ is the total number of items to choose from, and $k$ is the number of items to choose. In this case, we have a total of 10 people, and we need to split them into three committees of 3, 4, and 3 people. We can do this in the following steps: 1. Choose 3 people for the first committee: $C(10, 3)$ 2. Choose 4 people for the second committee from the remaining 7 people: $C(7, 4)$ 3. The remaining 3 people automatically form the third committee, so there's no need for a separate calculation. Now, let's calculate the combinations: 1. $C(10, 3) = \frac{10!}{3!(10-3)!} = \frac{10!}{3!7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$ 2. $C(7, 4) = \frac{7!}{4!(7-4)!} = \frac{7!}{4!3!} = \frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1} = 35$ Now, we need to multiply these two numbers to get the total number of ways to split the 10 people into 3 committees: $$120 \times 35 = 4200$$ So, there are 4200 different ways to split 10 people into 3 different committees of 3, 4, and 3 people, respectively. |
4200 |
| doubao | First, we select 3 people out of 10 for the first committee. This can be done in C(10,3) ways. Then, we select 4 people out of the remaining 7 for the second committee. This can be done in C(7,4) ways. Finally, the remaining 3 people form the third committee. The total number of ways is the product of these combinations: C(10,3) * C(7,4) = (10*9*8)/(3*2*1) * (7*6*5*4)/(4*3*2*1) = 120 * 35 = 4200. |
4200 |
| minimax | To solve this problem, we can use combinations. A combination is a selection of items from a larger pool where the order does not matter. The formula for combinations is: \[ C(n, k) = \frac{n!}{k!(n-k)!} \] where \( n \) is the total number of items, \( k \) is the number of items to choose, and \( ! \) denotes factorial. First, we select 3 people out of the 10 for the first committee. The number of ways to do this is: \[ C(10, 3) = \frac{10!}{3!(10-3)!} = \frac{10!}{3!7!} \] After selecting the first committee, we have 7 people left. Now we select 4 people out of the remaining 7 for the second committee: \[ C(7, 4) = \frac{7!}{4!(7-4)!} = \frac{7!}{4!3!} \] Finally, we have 3 people left for the third committee, and since we need all 3, there is only 1 way to choose them: \[ C(3, 3) = \frac{3!}{3!(3-3)!} = \frac{3!}{3!0!} = 1 \] Now, we multiply the number of ways to form each committee to get the total number of ways to form all three committees: \[ \text{Total ways} = C(10, 3) \times C(7, 4) \times C(3, 3) \] Let's calculate each term: \[ C(10, 3) = \frac{10!}{3!7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \] \[ C(7, 4) = \frac{7!}{4!3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \] \[ C(3, 3) = 1 \] Now multiply them together: \[ \text{Total ways} = 120 \times 35 \times 1 = 4200 \] So, there are 4,200 different ways to split the group of 10 people into committees of 3, 4, and 3 people. |
4200 |
| kimi | To solve this problem, we can use the concept of combinations in combinatorics. A combination is a selection of items without regard to the order in which they are selected. First, we need to choose 3 people out of 10 for the first committee. The number of ways to do this is given by the combination formula: \[ C(n, k) = \frac{n!}{k!(n-k)!} \] where \( n \) is the total number of items, \( k \) is the number of items to choose, and \( ! \) denotes factorial, which is the product of all positive integers up to that number. For the first committee: \[ C(10, 3) = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \] After choosing the first committee, we have 7 people left. We need to choose 4 people out of these 7 for the second committee: \[ C(7, 4) = \frac{7!}{4!(7-4)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \] After choosing the second committee, we have 3 people left, and they will automatically form the third committee. There is only 1 way to do this since there are no choices left. To find the total number of ways to split the group into the three committees, we multiply the number of ways to form each committee: \[ \text{Total ways} = C(10, 3) \times C(7, 4) \times C(3, 3) \] Since \( C(3, 3) = 1 \), the calculation simplifies to: \[ \text{Total ways} = 120 \times 35 \times 1 = 4200 \] So, there are 4200 different ways to split the group of 10 people into three committees of 3, 4, and 3 people, respectively. |
4200 |
| wenxinyiyan | To solve this problem, we need to determine the number of ways to split a group of 10 people into committees of 3, 4, and 3 people. This is a combination problem because the order of the people within each committee does not matter. We can use the combination formula to solve this problem. The combination formula is: C(n, k) = n! / (k!(n-k)!) where n is the total number of people and k is the number of people in the committee. For the first committee of 3 people, we have: C(10, 3) = 10! / (3!7!) = 120 For the second committee of 4 people, we have: C(7, 4) = 7! / (4!3!) = 35 For the third committee of 3 people, we have: C(3, 3) = 3! / (3!0!) = 1 Now, we multiply the combinations for each committee to get the total number of ways to split the group: 120 * 35 * 1 = 4200 So, there are 4200 ways to split a group of 10 people into committees of 3, 4, and 3 people. |
4200 |
| zhipu | To solve this problem, we need to use the concept of combinations, which is a way to calculate the number of possible groupings when the order does not matter. We have 10 people and we need to split them into three committees of 3, 4, and 3 people. The steps are as follows: 1. Choose 3 people out of 10 for the first committee. 2. Choose 4 people out of the remaining 7 for the second committee. 3. The last 3 people will automatically form the third committee. The number of ways to choose \( k \) people from a group of \( n \) people is given by the combination formula: \[ C(n, k) = \frac{n!}{k!(n - k)!} \] where \( n! \) is the factorial of \( n \) (the product of all positive integers up to \( n \)). Now let's calculate each step: 1. Choosing 3 out of 10: \[ C(10, 3) = \frac{10!}{3!(10 - 3)!} = \frac{10!}{3! \cdot 7!} \] 2. Choosing 4 out of the remaining 7: \[ C(7, 4) = \frac{7!}{4!(7 - 4)!} = \frac{7!}{4! \cdot 3!} \] 3. The last 3 people are chosen by default, so we don't need to calculate this step. Now we multiply the results of the first two steps to get the total number of ways to form the committees: \[ \text{Total ways} = C(10, 3) \times C(7, 4) \] Let's plug in the factorials and simplify: \[ C(10, 3) = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \] \[ C(7, 4) = \frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1} = 35 \] So the total number of ways is: \[ 120 \times 35 = 4200 \] Therefore, there are 4200 different ways to split 10 people into three committees of 3, 4, and 3 people, respectively. |
4200 |
| gpto1 | To solve this problem, we need to determine the number of ways to divide a group of 10 people into three distinct committees of sizes 3, 4, and 3, respectively. Since the committees are different (they are distinguishable), we use the concept of **multinomial coefficients** from combinatorics. ### Understanding the Problem: - **Total people**: 10 - **Committee sizes**: 3, 4, and 3 - **Committees are distinguishable**: This means that committee A (size 3) is different from committee B (size 4), which is different from committee C (size 3). The order and labeling of committees matter. ### Multinomial Coefficient Formula: The number of ways to partition a set of \( n \) distinct elements into \( r \) distinguishable groups of sizes \( n_1, n_2, \dots, n_r \) (where \( n_1 + n_2 + \dots + n_r = n \)) is given by the **multinomial coefficient**: \[ \text{Number of ways} = \dfrac{n!}{n_1! \times n_2! \times \dots \times n_r!} \] ### Applying the Formula: Let's assign: - \( n = 10 \) (total people) - \( n_1 = 3 \) (size of the first committee) - \( n_2 = 4 \) (size of the second committee) - \( n_3 = 3 \) (size of the third committee) Plugging these values into the multinomial coefficient formula: \[ \text{Number of ways} = \dfrac{10!}{3! \times 4! \times 3!} \] ### Calculating the Factorials: First, compute the factorials needed: - \( 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 3,628,800 \) - \( 3! = 3 \times 2 \times 1 = 6 \) - \( 4! = 4 \times 3 \times 2 \times 1 = 24 \) Now compute the denominator: \[ \text{Denominator} = 3! \times 4! \times 3! = 6 \times 24 \times 6 = 864 \] ### Performing the Division: Now divide the numerator by the denominator: \[ \text{Number of ways} = \dfrac{3,628,800}{864} \] To simplify the calculation, we can perform the division step by step. #### Simplifying the Division: 1. **Divide numerator and denominator by 16** (since 16 is a common factor of both 3,628,800 and 864): \[ \dfrac{3,628,800 \div 16}{864 \div 16} = \dfrac{226,800}{54} \] 2. **Divide numerator and denominator by 18** (since 18 divides both 226,800 and 54): \[ \dfrac{226,800 \div 18}{54 \div 18} = \dfrac{12,600}{3} \] 3. **Divide numerator and denominator**: \[ \dfrac{12,600}{3} = 4,200 \] Alternatively, you can perform the division directly: \[ \text{Number of ways} = \dfrac{3,628,800}{864} = 4,200 \] ### Conclusion: Therefore, there are **4,200 ways** to split the group of 10 people into three distinguishable committees of sizes 3, 4, and 3. **Answer:** **4,200** |
4200 |
| deepseekv3 | To determine the number of ways to split a group of 10 people into committees of sizes 3, 4, and 3, respectively, we can use the concept of **multinomial coefficients**. Here's a step-by-step solution: --- ### Step 1: Understand the Problem We need to divide 10 people into three groups: - Group A: 3 people - Group B: 4 people - Group C: 3 people The order of the groups matters only in terms of their sizes, not in terms of which group is labeled as A, B, or C. --- ### Step 2: Use the Multinomial Formula The number of ways to divide \( n \) distinct objects into groups of sizes \( k_1, k_2, \dots, k_r \) is given by the multinomial coefficient: \[ \frac{n!}{k_1! \cdot k_2! \cdot \dots \cdot k_r!} \] Here: - \( n = 10 \) (total number of people), - \( k_1 = 3 \) (size of Group A), - \( k_2 = 4 \) (size of Group B), - \( k_3 = 3 \) (size of Group C). Thus, the number of ways is: \[ \frac{10!}{3! \cdot 4! \cdot 3!} \] --- ### Step 3: Calculate the Factorials Compute the factorials: \[ 10! = 3628800, \quad 3! = 6, \quad 4! = 24 \] --- ### Step 4: Substitute and Simplify Substitute the factorials into the formula: \[ \frac{10!}{3! \cdot 4! \cdot 3!} = \frac{3628800}{6 \cdot 24 \cdot 6} \] Simplify the denominator: \[ 6 \cdot 24 \cdot 6 = 864 \] Now divide: \[ \frac{3628800}{864} = 4200 \] --- ### Step 5: Final Answer The number of ways to split the group is: \[ \boxed{4200} \] |
4200 |