| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | A The best answer is A |
A |
| deepseekr1 | The problem involves finding the sum of all integers \( k \) such that \( \binom{23}{4} + \binom{23}{5} = \binom{24}{k} \). Using Pascal's identity, \( \binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1} \), substituting \( n = 23 \) and \( r = 4 \), we get: \[ \binom{23}{4} + \binom{23}{5} = \binom{24}{5}. \] Due to the symmetry property of combinations (\( \binom{n}{k} = \binom{n}{n-k} \)), \( \binom{24}{5} = \binom{24}{19} \). Thus, \( k \) can be either 5 or 19. The sum of these values is \( 5 + 19 = 24 \). **Answer:** A. 24 |
A |
| gpto1 | To solve this problem, we need to find all integer values of \( k \) such that: \[ \binom{23}{4} + \binom{23}{5} = \binom{24}{k} \] First, let's recognize a key identity in combinatorics: \[ \binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k} \] Using this identity with \( n = 23 \) and \( k = 5 \): \[ \binom{23}{5} + \binom{23}{4} = \binom{24}{5} \] This simplifies our equation to: \[ \binom{24}{5} = \binom{24}{k} \] This implies that \( k \) must satisfy: \[ \binom{24}{k} = \binom{24}{5} \] Since the binomial coefficient satisfies \( \binom{n}{k} = \binom{n}{n - k} \), the values of \( k \) that satisfy the equation are: \[ k = 5 \quad \text{or} \quad k = 24 - 5 = 19 \] Therefore, the sum of all such integers \( k \) is: \[ 5 + 19 = 24 \] **Answer:** A. 24 |
A |
| GPT-4o | To find the sum of all integers \( k \) such that \( \binom{23}{4} + \binom{23}{5} = \binom{24}{k} \), we can use the Pascal's identity for binomial coefficients. Pascal's identity states that: \[ \binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}. \] In this case, we have \( n = 23 \) and \( r = 4 \). Applying Pascal's identity, we get: \[ \binom{23}{4} + \binom{23}{5} = \binom{24}{5}. \] So, \( \binom{24}{k} = \binom{24}{5} \). This implies that \( k = 5 \). Since there is only one \( k \) that satisfies the equation, the sum of all such \( k \) is simply \( 5 \). Therefore, the answer is: B. 5 |
A |
| llama3.1-70B-Instruct | A Pascal's Identity states that $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$. Thus, $\binom{23}{4} + \binom{23}{5} = \binom{24}{5}$. Since $\binom{24}{5} = \binom{24}{24-5} = \binom{24}{19}$, the sum of all integers $k$ such that $\binom{23}{4} + \binom{23}{5} = \binom{24}{k}$ is $5+19=24$. Final Answer: The final answer is 24. I hope it is correct. |
A |
| llama3.1-405B-Instruct | A The best answer is A. |
A |
| DeepSeek-Coder-V2 | A. 24 | A |
| Qwen2-72B-Instruct | A. 24 Explanation: The given equation is $\binom{23}{4} + \binom{23}{5} = \binom{24}{k}$. By Pascal's Identity, we know that $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$. Applying this identity to the given equation, we get $\binom{24}{5} = \binom{24}{k}$. The binomial coefficient $\binom{n}{r}$ is equal to $\binom{n}{n-r}$, so $\binom{24}{5} = \binom{24}{24-5} = \binom{24}{19}$. Therefore, the possible values of $k$ are 5 and 19. The sum of these values is $5 + 19 = 24$. |
A |
| baichuan4 | B | A |
| doubao | Using the combination formula $C_n^m=\frac{n!}{m!(n-m)!}$, we have: $\binom{23}{4}+\binom{23}{5}=\frac{23!}{4!(23-4)!}+\frac{23!}{5!(23-5)!}=\frac{23!}{4!19!}+\frac{23!}{5!18!}=\frac{23\times22\times21\times20}{4\times3\times2\times1}+\frac{23\times22\times21\times20\times19}{5\times4\times3\times2\times1}=\frac{23\times22\times21\times20}{24}+\frac{23\times22\times21\times20\times19}{120}=(23\times22\times21\times20)\times(\frac{1}{24}+\frac{19}{120})=(23\times22\times21\times20)\times(\frac{5}{120}+\frac{19}{120})=(23\times22\times21\times20)\times\frac{24}{120}=23\times11\times21\times2=253\times21$. And $\binom{24}{k}=\frac{24!}{k!(24-k)!}$. So we need to find $k$ such that $\frac{24!}{k!(24-k)!}=253\times21$. By checking the options, we find that when $k=5$, $\binom{24}{5}=\frac{24!}{5!(24-5)!}=\frac{24\times23\times22\times21\times20}{5\times4\times3\times2\times1}=253\times21$. So the answer is B. |
A |
| minimax | To solve this problem, we can use Pascal's rule, which states that for any non-negative integers $n$ and $r$: $$\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$$ Given the equation: $$\binom{23}{4} + \binom{23}{5} = \binom{24}{k}$$ We can apply Pascal's rule with $n = 23$ and $r = 4$: $$\binom{23}{4} + \binom{23}{5} = \binom{24}{5}$$ Therefore, we have: $$\binom{24}{k} = \binom{24}{5}$$ This implies that $k = 5$ because the binomial coefficient $\binom{n}{r}$ is equal to $\binom{n}{n-r}$, and there is only one value of $k$ that satisfies the equation. The answer is: B. 5 |
A |
| kimi | To solve this problem, we can use the identity for binomial coefficients that states: \[ \binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1} \] Given the equation: \[ \binom{23}{4} + \binom{23}{5} = \binom{24}{k} \] We can apply the identity with \( n = 23 \) and \( r = 4 \): \[ \binom{23}{4} + \binom{23}{5} = \binom{24}{5} \] This means that \( k = 5 \). Now we need to find the sum of all integers \( k \) that satisfy the equation. Since the binomial coefficient is symmetric, we also have: \[ \binom{24}{5} = \binom{24}{24-5} = \binom{24}{19} \] So, the values of \( k \) that satisfy the equation are \( k = 5 \) and \( k = 19 \). Adding these together: \[ k + (24 - k) = 5 + 19 = 24 \] Therefore, the sum of all integers \( k \) is 24. Answer: A. 24 |
A |
| wenxinyiyan | B Explanation: Using Pascal's Identity, which states that $\binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1}$, we can rewrite the given equation as follows: $\binom{23}{4} + \binom{23}{5} = \binom{24}{5}$ This is because when $n = 24$ and $k = 5$, Pascal's Identity gives: $\binom{24}{5} = \binom{23}{5} + \binom{23}{4}$ Since the equation $\binom{23}{4} + \binom{23}{5} = \binom{24}{k}$ holds true for $k = 5$, the sum of all integers $k$ that satisfy the equation is simply 5. |
A |
| zhipu | To solve the problem, we will use the properties of binomial coefficients. We start with the given equation: \[ \binom{23}{4} + \binom{23}{5} = \binom{24}{k} \] Using the property of binomial coefficients, specifically Pascal's identity, we know: \[ \binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1} \] In our case, \( n = 23 \) and \( r = 4 \). Applying Pascal's identity: \[ \binom{23}{4} + \binom{23}{5} = \binom{24}{5} \] This means: \[ \binom{24}{k} = \binom{24}{5} \] Since binomial coefficients are symmetric, we have: \[ \binom{24}{k} = \binom{24}{24-k} \] Therefore, \( k \) can be either 5 or \( 24 - 5 = 19 \). The sum of all integers \( k \) that satisfy the equation is: \[ 5 + 19 = 24 \] Thus, the correct answer is: \[ \boxed{24} \] |
A |
| gpto1 | To solve this problem, we need to find all integer values of \( k \) such that: \[ \binom{23}{4} + \binom{23}{5} = \binom{24}{k} \] First, let's recognize a key identity in combinatorics: \[ \binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k} \] Using this identity with \( n = 23 \) and \( k = 5 \): \[ \binom{23}{5} + \binom{23}{4} = \binom{24}{5} \] This simplifies our equation to: \[ \binom{24}{5} = \binom{24}{k} \] This implies that \( k \) must satisfy: \[ \binom{24}{k} = \binom{24}{5} \] Since the binomial coefficient satisfies \( \binom{n}{k} = \binom{n}{n - k} \), the values of \( k \) that satisfy the equation are: \[ k = 5 \quad \text{or} \quad k = 24 - 5 = 19 \] Therefore, the sum of all such integers \( k \) is: \[ 5 + 19 = 24 \] **Answer:** A. 24 |
A |
| deepseekv3 | To solve the problem, we start by simplifying the given equation: \[ \binom{23}{4} + \binom{23}{5} = \binom{24}{k} \] Using the **Pascal's Identity**, we know that: \[ \binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1} \] Applying this to the left-hand side of the equation with \( n = 23 \) and \( k = 4 \): \[ \binom{23}{4} + \binom{23}{5} = \binom{24}{5} \] Thus, the equation becomes: \[ \binom{24}{5} = \binom{24}{k} \] Since binomial coefficients are symmetric, \(\binom{24}{k} = \binom{24}{24 - k}\). Therefore, the possible values of \( k \) are: \[ k = 5 \quad \text{or} \quad k = 24 - 5 = 19 \] The sum of all possible integers \( k \) is: \[ 5 + 19 = 24 \] **Final Answer:** \boxed{A} |
A |