| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | D. 84 |
A |
| deepseekr1 | To determine the number of ways John can choose 4 marbles with exactly one being red, green, or blue, we break the problem into two parts: 1. **Choose 1 special marble (red, green, or blue):** There are 3 special marbles, so the number of ways to choose 1 is \( \binom{3}{1} = 3 \). 2. **Choose 3 non-special marbles from the remaining 9:** There are 9 non-special marbles. The number of ways to choose 3 is \( \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \). Multiplying these results gives the total number of valid combinations: \( 3 \times 84 = 252 \). **Verification via complementary counting:** - Total ways to choose 4 marbles: \( \binom{12}{4} = 495 \). - Subtract invalid cases (0, 2, or 3 special marbles): - 0 special: \( \binom{9}{4} = 126 \), - 2 special: \( \binom{3}{2} \times \binom{9}{2} = 3 \times 36 = 108 \), - 3 special: \( \binom{3}{3} \times \binom{9}{1} = 1 \times 9 = 9 \). - Total invalid: \( 126 + 108 + 9 = 243 \). - Valid combinations: \( 495 - 243 = 252 \). Thus, the answer is **A. 252**. **Answer:** A |
A |
| gpto1 | A | A |
| GPT-4o | To solve this problem, we need to determine the number of ways to choose 4 marbles from 12, such that exactly one of the chosen marbles is either red, green, or blue. Here is the step-by-step solution: 1. **Choose 1 out of the 3 colored marbles (red, green, or blue):** There are 3 choices for this step. 2. **Choose 3 marbles from the remaining 9 marbles:** After choosing one of the colored marbles, we need to choose 3 more marbles from the remaining 9 non-colored marbles. The number of ways to choose 3 marbles out of 9 is given by the combination formula \( \binom{9}{3} \). \[ \binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \] 3. **Multiply the number of choices:** Since there are 3 ways to choose the colored marble and 84 ways to choose the remaining 3 marbles, the total number of ways is: \[ 3 \times 84 = 252 \] Thus, the number of ways John can choose 4 marbles with exactly one being red, green, or blue is: Answer: A. 252 |
A |
| llama3.1-70B-Instruct | A |
A |
| llama3.1-405B-Instruct | C. 441 The best answer is C |
A |
| DeepSeek-Coder-V2 | B. 276 | A |
| Qwen2-72B-Instruct | B. 276 | A |
| baichuan4 | D | A |
| doubao | D. 84. First choose one of the red, green or blue in 3 ways. Then choose 3 marbles from the remaining 9 non-special marbles in C(9,3) ways. So the total number of ways is 3 * C(9,3) = 3 * 84 / 6 = 84. |
A |
| minimax | To solve this problem, we need to consider that exactly one of the chosen marbles must be red, green, or blue. There are 3 ways to choose which one of these three colors will be the single marble (either red, green, or blue). After choosing one of these colors, John needs to choose 3 more marbles from the remaining 9 marbles (since there are 12 marbles in total and 3 are already chosen by color). The number of ways to choose 3 marbles from 9 is given by 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, and \( ! \) denotes factorial. So, the number of ways to choose 3 marbles from the remaining 9 is: \[ \binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!} \] \[ = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} \] \[ = 3 \times 4 \times 7 \] \[ = 84 \] Since there are 3 ways to choose which color marble will be the single one, we multiply the number of ways to choose the other 3 marbles by 3: \[ 84 \times 3 = 252 \] Therefore, the number of ways John can choose 4 marbles with exactly one being red, green, or blue is 252. The correct answer is: A. 252 |
A |
| kimi | To solve this problem, we can break it down into two steps: 1. Choose the special marble (red, green, or blue). Since there are 3 special marbles, there are 3 ways to choose one of them. 2. Choose the remaining 3 marbles from the other 9 marbles (since we've already chosen one special marble, we are left with 12 - 3 = 9 marbles). The number of ways to choose 3 marbles from 9 is given 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. So, we have: \[ C(9, 3) = \frac{9!}{3!(9-3)!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \] Now, multiply the number of ways to choose the special marble by the number of ways to choose the remaining 3 marbles: \[ 3 \times 84 = 252 \] Therefore, the total number of ways John can choose 4 marbles with exactly one being red, green, or blue is 252. Answer: A. 252 |
A |
| wenxinyiyan | C Explanation: John has 12 marbles, including one red, one green, and one blue marble. He wants to choose 4 marbles, and exactly one of the chosen marbles should be red, green, or blue. We can break down the problem into three cases: 1. Choosing the red marble and 3 other non-red, non-green, non-blue marbles. 2. Choosing the green marble and 3 other non-red, non-green, non-blue marbles. 3. Choosing the blue marble and 3 other non-red, non-green, non-blue marbles. For each case, we use combination (C) to calculate the number of ways. 1. For the red marble case: C(1,1) × C(9,3) = 1 × 84 = 84 (Choose 1 red marble from 1 and choose 3 other marbles from the remaining 9) 2. For the green marble case: C(1,1) × C(9,3) = 1 × 84 = 84 (Choose 1 green marble from 1 and choose 3 other marbles from the remaining 9) 3. For the blue marble case: C(1,1) × C(9,3) = 1 × 84 = 84 (Choose 1 blue marble from 1 and choose 3 other marbles from the remaining 9) Total number of ways = 84 + 84 + 84 = 252 However, we need to consider that in each of these cases, the 3 other marbles chosen cannot be red, green, or blue. Since we have already accounted for the red, green, and blue marbles in each case, the 3 other marbles will always be from the remaining 9 non-red, non-green, non-blue marbles. Therefore, our initial calculation of 252 is correct, but we need to add the possibilities of choosing the same 3 non-colored marbles with different colored ones. For example, if we choose the red marble and then three other non-colored marbles (let's say marble A, B, and C), this is a valid choice. However, we can also choose the green marble with A, B, and C or the blue marble with A, B, and C. These are three separate valid choices, so we need to count them all. Since there are 84 ways to choose 3 non-colored marbles (C(9,3) = 84), and for each of these ways, we can choose either the red, green, or blue marble, we have a total of 84 * 3 = 252 valid choices. But wait, we actually counted each combination three times (once for each color). So, the final answer is not 252 but 252 + 2×84 = 420. However, this is still not among the options. What did we miss? The missing part is that when we choose the 3 non-colored marbles, they can also be chosen in different orders. For example, choosing A, B, C is different from choosing A, C, B. So, for each set of 3 non-colored marbles, there are actually 3! (3 factorial) = 6 ways to order them. Therefore, the final answer is 420 × 6 / 3! = 420 (since 6/3! = 1, the ordering doesn't matter in this case because we're only concerned with the combination of marbles, not their order). However, this is still incorrect because we overcounted. The correct approach is to simply consider the three cases (red, green, blue) separately and not worry about the order of the other three marbles since we're only interested in the combination, not the permutation. So, the final answer is indeed 252, which is the sum of the three cases: 84 (red) + 84 (green) + 84 (blue). But wait, there's a catch! The question asks for "exactly one of the chosen marbles is red, green, or blue." This means we actually need to exclude cases where more than one of the chosen marbles is red, green, or blue. Fortunately, in our case, this is not possible since there is only one red, one green, and one blue marble. So, we cannot choose more than one of each color. Therefore, the final |
A |
| zhipu | To solve this problem, we need to count the number of ways John can choose 4 marbles such that exactly one of them is either red, green, or blue. First, let's consider the number of ways to choose exactly one of the red, green, or blue marbles. There are 3 choices (red, green, or blue). Next, after choosing one of these 3 special marbles, John needs to choose 3 more marbles from the remaining 9 marbles (since there are 12 marbles in total and one is already chosen). The number of ways to choose 3 marbles from the remaining 9 is given by the combination formula: \[ \binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \] Since there are 3 different ways to choose the first special marble, the total number of ways to choose 4 marbles with exactly one being red, green, or blue is: \[ 3 \times 84 = 252 \] So, the correct answer is: A. 252 |
A |
| gpto1 | A | A |
| deepseekv3 | To solve the problem, let's break it down step by step. 1. **Total Marbles and Special Marbles**: - John has **12 marbles** in total. - Among them, **3 marbles** are special: **red**, **green**, and **blue**. - The remaining **9 marbles** are of other colors. 2. **Selection Criteria**: - John needs to choose **4 marbles**. - **Exactly one** of the chosen marbles must be **red**, **green**, or **blue**. 3. **Step-by-Step Calculation**: - **Choose 1 special marble** (red, green, or blue): There are **3** choices. - **Choose 3 marbles** from the remaining **9 marbles**: The number of ways to do this is given by the combination formula \( C(9, 3) \). \[ C(9, 3) = \frac{9!}{3! \times 6!} = 84 \] - **Total number of ways**: Multiply the number of ways to choose the special marble by the number of ways to choose the other marbles. \[ 3 \times 84 = 252 \] Therefore, the correct answer is: \boxed{A} |
A |