金声玉亮2.0模型评测报告

问题

There is a single choice question about college mathematics. Answer the question by replying A, B, C or D.
Question: A university’s mathematics department has 10 professors and will offer 20 different courses next semester. Each professor will be assigned to teach exactly 2 of the courses, and each course will have exactly one professor assigned to teach it. If any professor can be assigned to teach any course, how many different complete assignments of the 10 professors to the 20 courses are possible?
A. 20!/2^(10)
B. 10!/2^9
C. 10^(20) - 2^(10)
D. 10^(20) - 100
Answer:

答案

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 assigning 10 professors to teach 20 courses such that each professor teaches exactly 2 courses. To solve this, we consider two steps: partitioning the 20 courses into 10 pairs and assigning each pair to a distinct professor. 1. Partitioning Courses: The number of ways to partition 20 distinct courses into 10 unordered pairs is given by the formula \(\frac{20!}{2^{10} \cdot 10!}\). This accounts for dividing 20 items into pairs (divided by \(2^{10}\) for the indistinguishability of pairs and \(10!\) for the order of the pairs). 2. Assigning Pairs to Professors: Since the professors are distinct, we assign each of the 10 pairs to a professor, which can be done in \(10!\) ways. Combining these steps, the total number of assignments is: \[ \frac{20!}{2^{10} \cdot 10!} \times 10! = \frac{20!}{2^{10}}. \] This matches option A. Other options (B, C, D) involve incorrect formulas or irrelevant computations (e.g., subtracting terms from \(10^{20}\)), which do not align with the problem's constraints. Answer: A	A
gpto1	A	A
GPT-4o	A. 20!/2^(10)	A
llama3.1-70B-Instruct	A	A
llama3.1-405B-Instruct	The correct answer is A. 20!/2^(10). To see why, let's break down the problem step by step: 1. Each professor will be assigned to teach exactly 2 courses, so we need to choose 2 courses for each of the 10 professors. 2. For the first professor, there are 20 courses to choose from for the first course, and then 19 courses to choose from for the second course. This gives us 20 × 19 = 380 possible assignments for the first professor. 3. For the second professor, there are 18 courses left to choose from for the first course, and then 17 courses left to choose from for the second course. This gives us 18 × 17 = 306 possible assignments for the second professor. 4. We continue this process for all 10 professors, multiplying the number of possible assignments for each professor together: 380 × 306 × 234 × 182 × 140 × 108 × 84 × 66 × 52 × 40 = 20! / 2^10 The 2^10 term in the denominator comes from the fact that each professor is assigned to teach exactly 2 courses, so we need to divide by 2^10 to account for the fact that the order of the courses doesn't matter. Therefore, the correct answer is A. 20!/2^(10).	A
DeepSeek-Coder-V2	A. 20!/2^(10)	A
Qwen2-72B-Instruct	A. 20!/2^(10)	A
baichuan4	A	A
doubao	A The first course can be assigned to any of the 10 professors. The second course can be assigned to any of the remaining 9 professors. The third course can be assigned to any of the remaining 8 professors, and so on. So the total number of ways to assign the courses is 10×9×8×...×1 = 10!. But since each pair of courses can be assigned in either order, we need to divide by 2^10. So the answer is 20!/2^(10).	A
minimax	To solve this problem, we need to consider that each professor can teach any two of the twenty courses, and each course can be taught by any of the ten professors. We are looking for the number of ways to assign professors to courses such that each professor teaches exactly two courses and each course has exactly one professor. First, we choose 2 courses out of 20 for the first professor. This can be done in C(20, 2) ways. After assigning two courses to the first professor, we have 18 courses left. Next, we choose 2 courses out of the remaining 18 for the second professor. This can be done in C(18, 2) ways. We continue this process until all professors have been assigned two courses each. The total number of ways to assign courses to the professors is the product of the combinations for each professor: C(20, 2) * C(18, 2) * C(16, 2) * ... * C(4, 2) * C(2, 2) However, since the order in which we assign the courses to the professors does not matter (assigning professor A to courses 1 and 2 is the same as assigning professor A to courses 2 and 1), we have overcounted each assignment by a factor of 2! for each professor. Since there are 10 professors, we have overcounted by a factor of (2!)^10. Therefore, the correct number of different complete assignments is: C(20, 2) * C(18, 2) * ... * C(2, 2) / (2!)^10 This simplifies to: (20! / (2! * 18!)) * (18! / (2! * 16!)) * ... * (4! / (2! * 2!)) * (2! / (2! * 0!)) / (2!)^10 Notice that the factorials cancel out most of the terms, leaving us with: 20! / (2^10 * 10!) This matches the format of option A, which is the correct answer: A. 20!/2^(10)	A
kimi	A To solve this problem, we can think of it as a combinatorial problem. Each professor has to teach exactly 2 out of the 20 courses. We can use the combination formula to determine the number of ways to assign the courses to the professors. The number of ways to choose 2 courses out of 20 for the first professor is 20 choose 2, which is calculated as: \[ \binom{20}{2} = \frac{20!}{2!(20-2)!} = \frac{20 \times 19}{2 \times 1} = 190 \] After the first professor is assigned, there are 18 courses left. The second professor can choose from these 18 courses, so there are: \[ \binom{18}{2} = \frac{18!}{2!(18-2)!} = \frac{18 \times 17}{2 \times 1} = 153 \] This pattern continues until the last professor, who will have no choice but to teach the remaining 2 courses. The total number of ways to assign the courses is the product of all these combinations: \[ \binom{20}{2} \times \binom{18}{2} \times \binom{16}{2} \times \ldots \times \binom{2}{2} \] However, since each professor can teach any of the courses, we don't need to multiply by the number of professors (10) because the order in which we assign the professors doesn't matter. We are only concerned with how many ways we can assign the courses. The correct formula for the total number of assignments is: \[ \frac{20!}{2^{10}} \] This is because we are dividing the total permutations (20!) by the number of ways each professor can be assigned to two courses (2^10), since each professor has 2 choices for each of the 10 professors. So the correct answer is A. 20!/2^(10).	A
wenxinyiyan	To solve this problem, we need to determine the number of ways to assign 10 professors to 20 courses, with each professor teaching 2 courses and each course having one professor. 1. Choosing the First Professor's Courses: * The first professor can choose 2 courses out of 20. There are ${20 \choose 2} = \frac{20 imes 19}{2 imes 1}$ ways to do this. 2. Choosing the Second Professor's Courses: * After the first professor has chosen, there are 18 courses remaining. The second professor can choose 2 courses out of these 18, which can be done in ${18 \choose 2} = \frac{18 imes 17}{2 imes 1}$ ways. 3. Continuing the Process: * We continue this process for each of the remaining professors, choosing 2 courses from the remaining courses at each step. 4. Calculating the Total Number of Assignments: * The total number of assignments is the product of all these choices: ${20 \choose 2} imes {18 \choose 2} imes {16 \choose 2} imes \ldots imes {2 \choose 2}$. * This product can be simplified by noting that it is equivalent to $\frac{20!}{(2!)^{10}}$, where $20!$ represents the factorial of 20 (i.e., $20 imes 19 imes 18 imes \ldots imes 1$) and $2!$ represents the factorial of 2 (i.e., $2 imes 1$). The denominator $(2!)^{10}$ accounts for the fact that each professor teaches 2 courses. 5. Comparing with the Options: * The expression $\frac{20!}{(2!)^{10}}$ matches option A: $20!/2^{10}$ (since $2! = 2$ and $(2!)^{10} = 2^{10}$). Therefore, the correct answer is A.	A
zhipu	To solve this problem, we need to determine how many different ways we can assign 20 courses to 10 professors, where each professor teaches exactly 2 courses. 1. Step 1: Assign courses to professors: - We need to assign 20 different courses to 10 professors, with each professor teaching exactly 2 courses. - This can be viewed as dividing the 20 courses into 10 pairs, where each pair is taught by one professor. 2. Step 2: Calculate the number of ways to pair the courses: - First, we choose pairs of courses from the 20 courses. - The number of ways to choose the first pair is \(\binom{20}{2}\). - The number of ways to choose the second pair from the remaining 18 courses is \(\binom{18}{2}\), and so on. The total number of ways to pair up the 20 courses is: \[ \binom{20}{2} \times \binom{18}{2} \times \cdots \times \binom{4}{2} \times \binom{2}{2} \] This simplifies to: \[ \frac{20!}{(2!)^{10} \cdot 10!} \] 3. Step 3: Assign the pairs to professors: - Each pair of courses can be assigned to any of the 10 professors. - There are 10! ways to assign these 10 pairs to the 10 professors. 4. Step 4: Combine the two steps: - The total number of different complete assignments is the product of the number of ways to pair the courses and the number of ways to assign these pairs to the professors: \[ \frac{20!}{(2!)^{10} \cdot 10!} \times 10! \] Simplifying this, we get: \[ \frac{20!}{(2!)^{10}} \] 5. Step 5: Simplify further: - Note that \((2!)^{10} = 2^{10}\), so the expression becomes: \[ \frac{20!}{2^{10}} \] Therefore, the number of different complete assignments of the 10 professors to the 20 courses is: \[ \frac{20!}{2^{10}} \] So the correct answer is: A. 20!/2^(10)	A
gpto1	A	A
deepseekv3	To solve this problem, we need to determine the number of ways to assign 10 professors to 20 courses such that each professor teaches exactly 2 courses, and each course is taught by exactly one professor. 1. Step 1: Assign courses to professors - There are 20 courses, and each professor will teach exactly 2 courses. - The total number of ways to assign the courses to the professors is the number of ways to partition 20 courses into 10 groups of 2. 2. Step 2: Calculate the number of ways to partition the courses - The number of ways to partition 20 distinct items into 10 groups of 2 is given by: \[ \frac{20!}{(2!)^{10}} \] - Here, \(20!\) is the number of ways to arrange 20 courses, and \((2!)^{10}\) accounts for the fact that the order within each pair of courses assigned to a professor doesn't matter. 3. Step 3: Simplify the expression - The expression simplifies to: \[ \frac{20!}{2^{10}} \] - This is because \(2! = 2\), so \((2!)^{10} = 2^{10}\). 4. Step 4: Match with the given options - The correct answer matches option A: \[ \frac{20!}{2^{10}} \] Answer: A	A

模型

模型答案

正确答案

金声玉亮2.0

A
The best answer is A

deepseekr1

The problem involves assigning 10 professors to teach 20 courses such that each professor teaches exactly 2 courses. To solve this, we consider two steps: partitioning the 20 courses into 10 pairs and assigning each pair to a distinct professor.

1. **Partitioning Courses**: The number of ways to partition 20 distinct courses into 10 unordered pairs is given by the formula $\frac{20!}{2^{10} \cdot 10!}$. This accounts for dividing 20 items into pairs (divided by $2^{10}$ for the indistinguishability of pairs and $10!$ for the order of the pairs).

2. **Assigning Pairs to Professors**: Since the professors are distinct, we assign each of the 10 pairs to a professor, which can be done in $10!$ ways.

Combining these steps, the total number of assignments is:
\[
\frac{20!}{2^{10} \cdot 10!} \times 10! = \frac{20!}{2^{10}}.
\]

This matches **option A**. Other options (B, C, D) involve incorrect formulas or irrelevant computations (e.g., subtracting terms from $10^{20}$), which do not align with the problem's constraints.

**Answer:** A

gpto1

GPT-4o

A. 20!/2^(10)

llama3.1-70B-Instruct

llama3.1-405B-Instruct

The correct answer is A. 20!/2^(10).

To see why, let's break down the problem step by step:

1. Each professor will be assigned to teach exactly 2 courses, so we need to choose 2 courses for each of the 10 professors.
2. For the first professor, there are 20 courses to choose from for the first course, and then 19 courses to choose from for the second course. This gives us 20 × 19 = 380 possible assignments for the first professor.
3. For the second professor, there are 18 courses left to choose from for the first course, and then 17 courses left to choose from for the second course. This gives us 18 × 17 = 306 possible assignments for the second professor.
4. We continue this process for all 10 professors, multiplying the number of possible assignments for each professor together:

380 × 306 × 234 × 182 × 140 × 108 × 84 × 66 × 52 × 40 = 20! / 2^10

The 2^10 term in the denominator comes from the fact that each professor is assigned to teach exactly 2 courses, so we need to divide by 2^10 to account for the fact that the order of the courses doesn't matter.

Therefore, the correct answer is A. 20!/2^(10).

DeepSeek-Coder-V2

A. 20!/2^(10)

Qwen2-72B-Instruct

A. 20!/2^(10)

baichuan4

doubao

A

The first course can be assigned to any of the 10 professors. The second course can be assigned to any of the remaining 9 professors. The third course can be assigned to any of the remaining 8 professors, and so on. So the total number of ways to assign the courses is 10×9×8×...×1 = 10!. But since each pair of courses can be assigned in either order, we need to divide by 2^10. So the answer is 20!/2^(10).

minimax

To solve this problem, we need to consider that each professor can teach any two of the twenty courses, and each course can be taught by any of the ten professors. We are looking for the number of ways to assign professors to courses such that each professor teaches exactly two courses and each course has exactly one professor.

First, we choose 2 courses out of 20 for the first professor. This can be done in C(20, 2) ways. After assigning two courses to the first professor, we have 18 courses left.

Next, we choose 2 courses out of the remaining 18 for the second professor. This can be done in C(18, 2) ways. We continue this process until all professors have been assigned two courses each.

The total number of ways to assign courses to the professors is the product of the combinations for each professor:

C(20, 2) * C(18, 2) * C(16, 2) * ... * C(4, 2) * C(2, 2)

However, since the order in which we assign the courses to the professors does not matter (assigning professor A to courses 1 and 2 is the same as assigning professor A to courses 2 and 1), we have overcounted each assignment by a factor of 2! for each professor. Since there are 10 professors, we have overcounted by a factor of (2!)^10.

Therefore, the correct number of different complete assignments is:

C(20, 2) * C(18, 2) * ... * C(2, 2) / (2!)^10

This simplifies to:

(20! / (2! * 18!)) * (18! / (2! * 16!)) * ... * (4! / (2! * 2!)) * (2! / (2! * 0!)) / (2!)^10

Notice that the factorials cancel out most of the terms, leaving us with:

20! / (2^10 * 10!)

This matches the format of option A, which is the correct answer:

A. 20!/2^(10)

kimi

A

To solve this problem, we can think of it as a combinatorial problem. Each professor has to teach exactly 2 out of the 20 courses. We can use the combination formula to determine the number of ways to assign the courses to the professors.

The number of ways to choose 2 courses out of 20 for the first professor is 20 choose 2, which is calculated as:

\[
\binom{20}{2} = \frac{20!}{2!(20-2)!} = \frac{20 \times 19}{2 \times 1} = 190
\]

After the first professor is assigned, there are 18 courses left. The second professor can choose from these 18 courses, so there are:

\[
\binom{18}{2} = \frac{18!}{2!(18-2)!} = \frac{18 \times 17}{2 \times 1} = 153
\]

This pattern continues until the last professor, who will have no choice but to teach the remaining 2 courses.

The total number of ways to assign the courses is the product of all these combinations:

\[
\binom{20}{2} \times \binom{18}{2} \times \binom{16}{2} \times \ldots \times \binom{2}{2}
\]

However, since each professor can teach any of the courses, we don't need to multiply by the number of professors (10) because the order in which we assign the professors doesn't matter. We are only concerned with how many ways we can assign the courses.

The correct formula for the total number of assignments is:

\[
\frac{20!}{2^{10}}
\]

This is because we are dividing the total permutations (20!) by the number of ways each professor can be assigned to two courses (2^10), since each professor has 2 choices for each of the 10 professors.

So the correct answer is A. 20!/2^(10).

wenxinyiyan

To solve this problem, we need to determine the number of ways to assign 10 professors to 20 courses, with each professor teaching 2 courses and each course having one professor.

1. **Choosing the First Professor's Courses**:

* The first professor can choose 2 courses out of 20. There are ${20 \choose 2} = \frac{20 imes 19}{2 imes 1}$ ways to do this.
2. **Choosing the Second Professor's Courses**:

* After the first professor has chosen, there are 18 courses remaining. The second professor can choose 2 courses out of these 18, which can be done in ${18 \choose 2} = \frac{18 imes 17}{2 imes 1}$ ways.
3. **Continuing the Process**:

* We continue this process for each of the remaining professors, choosing 2 courses from the remaining courses at each step.
4. **Calculating the Total Number of Assignments**:

* The total number of assignments is the product of all these choices: ${20 \choose 2} imes {18 \choose 2} imes {16 \choose 2} imes \ldots imes {2 \choose 2}$.
* This product can be simplified by noting that it is equivalent to $\frac{20!}{(2!)^{10}}$, where $20!$ represents the factorial of 20 (i.e., $20 imes 19 imes 18 imes \ldots imes 1$) and $2!$ represents the factorial of 2 (i.e., $2 imes 1$). The denominator $(2!)^{10}$ accounts for the fact that each professor teaches 2 courses.
5. **Comparing with the Options**:

* The expression $\frac{20!}{(2!)^{10}}$ matches option A: $20!/2^{10}$ (since $2! = 2$ and $(2!)^{10} = 2^{10}$).

Therefore, the correct answer is **A**.

zhipu

To solve this problem, we need to determine how many different ways we can assign 20 courses to 10 professors, where each professor teaches exactly 2 courses.

1. **Step 1: Assign courses to professors:**
- We need to assign 20 different courses to 10 professors, with each professor teaching exactly 2 courses.
- This can be viewed as dividing the 20 courses into 10 pairs, where each pair is taught by one professor.

2. **Step 2: Calculate the number of ways to pair the courses:**
- First, we choose pairs of courses from the 20 courses.
- The number of ways to choose the first pair is $\binom{20}{2}$.
- The number of ways to choose the second pair from the remaining 18 courses is $\binom{18}{2}$, and so on.

The total number of ways to pair up the 20 courses is:
\[
\binom{20}{2} \times \binom{18}{2} \times \cdots \times \binom{4}{2} \times \binom{2}{2}
\]

This simplifies to:
\[
\frac{20!}{(2!)^{10} \cdot 10!}
\]

3. **Step 3: Assign the pairs to professors:**
- Each pair of courses can be assigned to any of the 10 professors.
- There are 10! ways to assign these 10 pairs to the 10 professors.

4. **Step 4: Combine the two steps:**
- The total number of different complete assignments is the product of the number of ways to pair the courses and the number of ways to assign these pairs to the professors:
\[
\frac{20!}{(2!)^{10} \cdot 10!} \times 10!
\]
Simplifying this, we get:
\[
\frac{20!}{(2!)^{10}}
\]

5. **Step 5: Simplify further:**
- Note that $(2!)^{10} = 2^{10}$, so the expression becomes:
\[
\frac{20!}{2^{10}}
\]

Therefore, the number of different complete assignments of the 10 professors to the 20 courses is:
\[
\frac{20!}{2^{10}}
\]

So the correct answer is:
**A. 20!/2^(10)**

gpto1

deepseekv3

To solve this problem, we need to determine the number of ways to assign 10 professors to 20 courses such that each professor teaches exactly 2 courses, and each course is taught by exactly one professor.

1. **Step 1: Assign courses to professors**
- There are 20 courses, and each professor will teach exactly 2 courses.
- The total number of ways to assign the courses to the professors is the number of ways to partition 20 courses into 10 groups of 2.

2. **Step 2: Calculate the number of ways to partition the courses**
- The number of ways to partition 20 distinct items into 10 groups of 2 is given by:
\[
\frac{20!}{(2!)^{10}}
\]
- Here, $20!$ is the number of ways to arrange 20 courses, and $(2!)^{10}$ accounts for the fact that the order within each pair of courses assigned to a professor doesn't matter.

3. **Step 3: Simplify the expression**
- The expression simplifies to:
\[
\frac{20!}{2^{10}}
\]
- This is because $2! = 2$, so $(2!)^{10} = 2^{10}$.

4. **Step 4: Match with the given options**
- The correct answer matches option A:
\[
\frac{20!}{2^{10}}
\]

**Answer: A**