| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve this problem, we can use the combination formula which is given by: C(n, r) = n! / [r!(n-r)!] where n is the total number of items, r is the number of items to be selected, and "!" denotes factorial, which means the product of all positive integers up to that number. First, we need to select 2 men out of 6. The number of ways to do this is: C(6, 2) = 6! / [2!(6-2)!] = 6! / (2! * 4!) = (6 * 5) / (2 * 1) = 15 Next, we need to select 3 women out of 8. The number of ways to do this is: C(8, 3) = 8! / [3!(8-3)!] = 8! / (3! * 5!) = (8 * 7 * 6) / (3 * 2 * 1) = 56 Finally, we need to multiply the number of ways to select the men by the number of ways to select the women, since these are independent events. Therefore, the total number of ways to select a committee of 2 men and 3 women is: 15 * 56 = 840 Therefore, a committee of 2 men and 3 women can be selected from a group of 6 men and 8 women in 840 ways. |
840 |
| deepseekr1 | To determine the number of ways to select a committee of 2 men and 3 women from 6 men and 8 women, we use combinations since the order of selection does not matter. 1. **Calculate the number of ways to choose 2 men from 6:** \[ \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \] 2. **Calculate the number of ways to choose 3 women from 8:** \[ \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \] 3. **Multiply the two results to find the total number of committees:** \[ 15 \times 56 = 840 \] **Answer:** There are \(\boxed{840}\) ways to form the committee. |
840 |
| gpto1 | To find out how many ways a committee of 2 men and 3 women can be selected from a group of 6 men and 8 women, we'll use combinations from combinatorics. **Step 1: Choose 2 men out of 6 men** The number of ways to choose 2 men from 6 is given by the combination formula: \[ C(n, k) = \frac{n!}{k!(n - k)!} \] So, \[ C(6, 2) = \frac{6!}{2!(6 - 2)!} = \frac{6!}{2!4!} = \frac{6 \times 5 \times 4!}{2 \times 1 \times 4!} = \frac{30}{2} = 15 \] There are **15** ways to choose 2 men. **Step 2: Choose 3 women out of 8 women** Similarly, \[ C(8, 3) = \frac{8!}{3!(8 - 3)!} = \frac{8!}{3!5!} = \frac{8 \times 7 \times 6 \times 5!}{6 \times 1 \times 5!} = \frac{336}{6} = 56 \] There are **56** ways to choose 3 women. **Step 3: Multiply the number of ways** Since the selections of men and women are independent events, we multiply the number of ways together: \[ 15 \text{ ways (men)} \times 56 \text{ ways (women)} = 840 \text{ total ways} \] **Answer:** 840 |
840 |
| GPT-4o | To solve the problem of selecting a committee of 2 men and 3 women from a group of 6 men and 8 women, we can use the concept of combinations. The number of ways to choose a subset of a certain size from a larger set is given by the combination formula: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] where \( n \) is the total number of items, \( k \) is the number of items to choose, and \( ! \) denotes the factorial of a number. 1. **Selecting 2 men from 6 men:** The number of ways to choose 2 men out of 6 is given by: \[ \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6!}{2! \cdot 4!} \] Calculating the factorials, we get: \[ 6! = 720, \quad 2! = 2, \quad 4! = 24 \] So, \[ \binom{6}{2} = \frac{720}{2 \cdot 24} = \frac{720}{48} = 15 \] 2. **Selecting 3 women from 8 women:** The number of ways to choose 3 women out of 8 is given by: \[ \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8!}{3! \cdot 5!} \] Calculating the factorials, we get: \[ 8! = 40320, \quad 3! = 6, \quad 5! = 120 \] So, \[ \binom{8}{3} = \frac{40320}{6 \cdot 120} = \frac{40320}{720} = 56 \] 3. **Combining the selections:** Since the selection of men and women are independent events, we multiply the number of ways to choose the men by the number of ways to choose the women: \[ \binom{6}{2} \times \binom{8}{3} = 15 \times 56 = 840 \] Therefore, the number of ways to select a committee of 2 men and 3 women from a group of 6 men and 8 women is \( \boxed{840} \). |
840 |
| llama3.1-70B-Instruct | To select a committee of 2 men and 3 women from a group of 6 men and 8 women, we can use the concept of combinations. **Step 1: Select 2 men from 6 men** The number of ways to select 2 men from 6 men is given by the combination formula: C(6, 2) = 6! / (2! × (6-2)!) = 6! / (2! × 4!) = (6 × 5) / (2 × 1) = 15 **Step 2: Select 3 women from 8 women** The number of ways to select 3 women from 8 women is given by the combination formula: C(8, 3) = 8! / (3! × (8-3)!) = 8! / (3! × 5!) = (8 × 7 × 6) / (3 × 2 × 1) = 56 **Step 3: Multiply the number of ways to select men and women** Since the selection of men and women is independent, we multiply the number of ways to select 2 men and 3 women: 15 × 56 = 840 Therefore, there are **840** ways to select a committee of 2 men and 3 women from a group of 6 men and 8 women. |
840 |
| llama3.1-405B-Instruct | To select a committee of 2 men and 3 women from a group of 6 men and 8 women, we can use the concept of combinations. First, we select 2 men from 6 men. This can be done in 6C2 ways, where 6C2 is the number of combinations of 6 items taken 2 at a time. 6C2 = 6! / (2! \* (6-2)!) = 6! / (2! \* 4!) = (6 \* 5) / (2 \* 1) = 15 Next, we select 3 women from 8 women. This can be done in 8C3 ways, where 8C3 is the number of combinations of 8 items taken 3 at a time. 8C3 = 8! / (3! \* (8-3)!) = 8! / (3! \* 5!) = (8 \* 7 \* 6) / (3 \* 2 \* 1) = 56 Now, we need to find the total number of ways to select 2 men and 3 women. Since the selection of men and women are independent, we can multiply the number of ways to select men and women. Total number of ways = 6C2 \* 8C3 = 15 \* 56 = 840 Therefore, there are 840 ways to select a committee of 2 men and 3 women from a group of 6 men and 8 women. |
840 |
| DeepSeek-Coder-V2 | To solve the problem of selecting a committee of 2 men and 3 women from a group of 6 men and 8 women, we need to use combinations. Combinations are used when the order of selection does not matter. Step-by-step solution: 1. **Calculate the number of ways to choose 2 men from 6 men:** The formula for combinations is given by: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] Here, \( n = 6 \) and \( k = 2 \): \[ \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6!}{2!4!} \] Simplify the factorials: \[ \binom{6}{2} = \frac{6 \times 5 \times 4!}{2 \times 1 \times 4!} = \frac{6 \times 5}{2 \times 1} = \frac{30}{2} = 15 \] So, there are 15 ways to choose 2 men from 6 men. 2. **Calculate the number of ways to choose 3 women from 8 women:** Here, \( n = 8 \) and \( k = 3 \): \[ \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!} \] Simplify the factorials: \[ \binom{8}{3} = \frac{8 \times 7 \times 6 \times 5!}{3 \times 2 \times 1 \times 5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56 \] So, there are 56 ways to choose 3 women from 8 women. 3. **Calculate the total number of ways to form the committee:** Since the selection of men and women are independent events, we multiply the number of ways to choose the men by the number of ways to choose the women: \[ \text{Total number of ways} = \binom{6}{2} \times \binom{8}{3} = 15 \times 56 = 840 \] Therefore, the number of ways to select a committee of 2 men and 3 women from a group of 6 men and 8 women is: \[ \boxed{840} \] |
840 |
| Qwen2-72B-Instruct | To solve this problem, we can use the combination formula which is given by: C(n, r) = n! / [r!(n-r)!] where n is the total number of items, r is the number of items to be selected, and "!" denotes factorial, which means the product of all positive integers up to that number. First, we need to select 2 men out of 6. The number of ways to do this is: C(6, 2) = 6! / [2!(6-2)!] = 6! / (2! * 4!) = (6 * 5) / (2 * 1) = 15 Next, we need to select 3 women out of 8. The number of ways to do this is: C(8, 3) = 8! / [3!(8-3)!] = 8! / (3! * 5!) = (8 * 7 * 6) / (3 * 2 * 1) = 56 Finally, we need to multiply the number of ways to select the men by the number of ways to select the women, since these are independent events. Therefore, the total number of ways to select a committee of 2 men and 3 women is: 15 * 56 = 840 Therefore, a committee of 2 men and 3 women can be selected from a group of 6 men and 8 women in 840 ways. |
840 |
| baichuan4 | To solve this problem, we can use the concept of combinations. A combination is a way of selecting items from a set where the order does not matter. In this case, we need to select 2 men and 3 women from a group of 6 men and 8 women. First, let's calculate the number of ways to select 2 men from 6 men. This can be done using the formula for combinations: C(n, k) = n! / (k! * (n - k)!) where n is the total number of items and k is the number of items to select. For selecting 2 men from 6 men, we have: C(6, 2) = 6! / (2! * (6 - 2)!) = (6 * 5) / (2 * 1) = 15 ways Next, let's calculate the number of ways to select 3 women from 8 women: C(8, 3) = 8! / (3! * (8 - 3)!) = (8 * 7 * 6) / (3 * 2 * 1) = 56 ways Now, since we need to select both 2 men and 3 women, we can multiply the number of ways to select men and women: 15 (ways to select men) * 56 (ways to select women) = 840 ways So, there are 840 different ways to select a committee of 2 men and 3 women from a group of 6 men and 8 women. |
840 |
| doubao | The number of ways to select 2 men from 6 men is 6C2 = (6*5)/(2*1) = 15. The number of ways to select 3 women from 8 women is 8C3 = (8*7*6)/(3*2*1) = 56. To find the total number of ways to form the committee, we multiply the number of ways of selecting men and women: 15 * 56 = 840. So, the committee can be formed in 840 ways. |
840 |
| minimax | To solve this problem, we can use combinations, since the order in which we select the committee members does not matter. First, we need to select 2 men out of 6. The number of ways to do this is given by the combination formula: C(n, k) = n! / (k! * (n - k)!) where n is the total number of items to choose from, k is the number of items to choose, and "!" denotes factorial. For the men, we have: C(6, 2) = 6! / (2! * (6 - 2)!) = (6 * 5 * 4 * 3 * 2 * 1) / ((2 * 1) * (4 * 3 * 2 * 1)) = (6 * 5) / (2 * 1) = 30 / 2 = 15 ways Next, we need to select 3 women out of 8. Using the combination formula again: C(8, 3) = 8! / (3! * (8 - 3)!) = (8 * 7 * 6 * 5 * 4 * 3 * 2 * 1) / ((3 * 2 * 1) * (5 * 4 * 3 * 2 * 1)) = (8 * 7 * 6) / (3 * 2 * 1) = (8 * 7 * 2) = 112 ways Now, to form the committee, we need to select both the men and the women. Since these are independent choices, we multiply the number of ways to choose the men by the number of ways to choose the women: Total ways = Ways to choose men * Ways to choose women = 15 * 112 = 1680 ways Therefore, there are 1680 different ways to select a committee of 2 men and 3 women from a group of 6 men and 8 women. |
840 |
| kimi | To solve this problem, we can use the concept of combinations. A combination is a selection of items from a larger set, where the order of the items does not matter. The number of ways to select 2 men from a group of 6 men is calculated by the combination formula: \[ C(n, k) = \frac{n!}{k!(n-k)!} \] where \( n \) is the total number of items to choose from, \( k \) is the number of items to choose, and \( ! \) denotes factorial, which is the product of all positive integers up to that number. For 2 men from 6 men: \[ C(6, 2) = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \] Similarly, the number of ways to select 3 women from a group of 8 women is: \[ C(8, 3) = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \] Now, to find the total number of ways to select a committee of 2 men and 3 women, we multiply the number of ways to choose the men by the number of ways to choose the women: \[ \text{Total ways} = C(6, 2) \times C(8, 3) = 15 \times 56 = 840 \] So, there are 840 different ways to select a committee of 2 men and 3 women from a group of 6 men and 8 women. |
840 |
| wenxinyiyan | To solve this problem, we can use the combination formula, which is denoted as "C(n, k)" or "n choose k". This formula calculates the number of ways to choose k items from a set of n items, without regard to order. In this case, we have two separate selections to make: selecting 2 men from 6 men, and selecting 3 women from 8 women. 1. Selecting 2 men from 6 men: We use the combination formula C(6, 2) to calculate the number of ways to choose 2 men from 6 men. C(6, 2) = 6! / (2! * (6-2)!) = 6*5 / (2*1) = 15 So, there are 15 ways to select 2 men from 6 men. 2. Selecting 3 women from 8 women: Similarly, we use the combination formula C(8, 3) to calculate the number of ways to choose 3 women from 8 women. C(8, 3) = 8! / (3! * (8-3)!) = (8*7*6) / (3*2*1) = 56 Thus, there are 56 ways to select 3 women from 8 women. To find the total number of ways to select a committee of 2 men and 3 women, we multiply the number of ways to select the men by the number of ways to select the women: Total number of ways = 15 (ways to select men) * 56 (ways to select women) = 840 Therefore, there are 840 ways to select a committee of 2 men and 3 women from a group of 6 men and 8 women. |
840 |
| zhipu | To solve this problem, we can use the concept of combinations, which is a way to calculate the number of possible groupings when the order does not matter. The number of ways to choose a committee of 2 men from a group of 6 men can be calculated using the combination formula: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] where: - \( n \) is the total number of items to choose from, - \( k \) is the number of items to choose, - \( n! \) denotes the factorial of \( n \), which is the product of all positive integers up to \( n \). For the men: \[ \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \] So there are 15 ways to choose 2 men from 6. For the women: \[ \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \] So there are 56 ways to choose 3 women from 8. To find the total number of ways to form a committee of 2 men and 3 women, we multiply the number of ways to choose the men by the number of ways to choose the women: \[ \text{Total ways} = \binom{6}{2} \times \binom{8}{3} = 15 \times 56 = 840 \] Therefore, there are 840 different ways to select a committee of 2 men and 3 women from a group of 6 men and 8 women. |
840 |
| gpto1 | To find out how many ways a committee of 2 men and 3 women can be selected from a group of 6 men and 8 women, we'll use combinations from combinatorics. **Step 1: Choose 2 men out of 6 men** The number of ways to choose 2 men from 6 is given by the combination formula: \[ C(n, k) = \frac{n!}{k!(n - k)!} \] So, \[ C(6, 2) = \frac{6!}{2!(6 - 2)!} = \frac{6!}{2!4!} = \frac{6 \times 5 \times 4!}{2 \times 1 \times 4!} = \frac{30}{2} = 15 \] There are **15** ways to choose 2 men. **Step 2: Choose 3 women out of 8 women** Similarly, \[ C(8, 3) = \frac{8!}{3!(8 - 3)!} = \frac{8!}{3!5!} = \frac{8 \times 7 \times 6 \times 5!}{6 \times 1 \times 5!} = \frac{336}{6} = 56 \] There are **56** ways to choose 3 women. **Step 3: Multiply the number of ways** Since the selections of men and women are independent events, we multiply the number of ways together: \[ 15 \text{ ways (men)} \times 56 \text{ ways (women)} = 840 \text{ total ways} \] **Answer:** 840 |
840 |
| deepseekv3 | To determine the number of ways to select a committee of 2 men and 3 women from a group of 6 men and 8 women, we can use combinations. Combinations are used when the order of selection does not matter. ### Step 1: Select 2 Men from 6 The number of ways to choose 2 men out of 6 is given by the combination formula: \[ \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \] ### Step 2: Select 3 Women from 8 The number of ways to choose 3 women out of 8 is given by the combination formula: \[ \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \] ### Step 3: Multiply the Two Results Since the selection of men and women are independent events, we multiply the number of ways to choose the men by the number of ways to choose the women: \[ 15 \times 56 = 840 \] ### Final Answer: There are **840** ways to select a committee of 2 men and 3 women from a group of 6 men and 8 women. |
840 |