Problem | Model | Prompt |
---|---|---|

P_Combinatorics_8 | GPT-3.5 | No C w/ H |

Ground Truth Answer: 634 |

```
import champ_dataset # "pip install champ-dataset" if necessary
dataset = champ_dataset.load('v0')
problem = dataset['P_Combinatorics_8']
messages = problem.conversations['GPT-3.5|No C w/ H'].messages
for message in messages:
print(f'Role: {message.role}')
print(f'Text: {message.text}')
print(f'Error: {message.error}')
print('-----------------------')
```

`You are an expert on mathematics.`

```
Solve the following problem. Make sure to show your work before giving the final answer.
How many strings of length 5 and consisting solely of the digits 1, 2, 3 and 4 have the property that any 1 and 2 are never neighbors (i.e., the string does not contain "12" or "21" as substrings).
You may find the following information useful:
1. Let a_n, b_n, c_n, d_n be the numbers of length-n strings that end with 1, 2, 3, 4 respectively.
2. Express a_(n+1), b_(n+1), c_(n+1), d_(n+1) in terms of a_n, b_n, c_n, d_n.
```

```
To solve this problem, we can use the principle of inclusion-exclusion.
Let's consider the possible cases for the last digit of the string:
Case 1: The last digit is 1.
In this case, the previous digit cannot be 2. So, we can have any of the digits 1, 3, or 4 before the last digit. Therefore, the number of strings of length n+1 that end with 1 is equal to the total number of strings of length n that end with 1, 3, or 4. This can be represented as a_(n+1) = a_n + c_n + d_n.
Case 2: The last digit is 2.
Similar to Case 1, the previous digit cannot be 1. So, we can have any of the digits 2, 3, or 4 before the last digit. Therefore, the number of strings of length n+1 that end with 2 is equal to the total number of strings of length n that end with 2, 3, or 4. This can be represented as b_(n+1) = b_n + c_n + d_n.
Case 3: The last digit is 3.
In this case, the previous digit can be any of the digits 1, 2, 3, or 4. Therefore, the number of strings of length n+1 that end with 3 is equal to the total number of strings of length n. This can be represented as c_(n+1) = a_n + b_n + c_n + d_n.
Case 4: The last digit is 4.
Similar to Case 3, the previous digit can be any of the digits 1, 2, 3, or 4. Therefore, the number of strings of length n+1 that end with 4 is equal to the total number of strings of length n. This can be represented as d_(n+1) = a_n + b_n + c_n + d_n.
Now, let's consider the base cases:
For n = 1, we have a_1 = 1, b_1 = 1, c_1 = 1, d_1 = 1 since there is only one digit in the string.
Using the recurrence relations derived above, we can calculate the values of a_n, b_n, c_n, d_n for n = 2, 3, 4, 5.
For n = 2:
a_2 = a_1 + c_1 + d_1 = 1 + 1 + 1 = 3
b_2 = b_1 + c_1 + d_1 = 1 + 1 + 1 = 3
c_2 = a_1 + b_1 + c_1 + d_1 = 1 + 1 + 1 + 1 = 4
d_2 = a_1 + b_1 + c_1 + d_1 = 1 + 1 + 1 + 1 = 4
For n = 3:
a_3 = a_2 + c_2 + d_2 = 3 + 4 + 4 = 11
b_3 = b_2 + c_2 + d_2 = 3 + 4 + 4 = 11
c_3 = a_2 + b_2 + c_2 + d_2 = 3 + 3 + 4 + 4 = 14
d_3 = a_2 + b_2 + c_2 + d_2 = 3 + 3 + 4 + 4 = 14
For n = 4:
a_4 = a_3 + c_3 + d_3 = 11 + 14 + 14 = 39
b_4 = b_3 + c_3 + d_3 = 11 + 14 + 14 = 39
c_4 = a_3 + b_3 + c_3 + d_3 = 11 + 11 + 14 + 14 = 50
d_4 = a_3 + b_3 + c_3 + d_3 = 11 + 11 + 14 + 14 = 50
For n = 5:
a_5 = a_4 + c_4 + d_4 = 39 + 50 + 50 = 139
b_5 = b_4 + c_4 + d_4 = 39 + 50 + 50 = 139
c_5 = a_4 + b_4 + c_4 + d_4 = 39 + 39 + 50 + 50 = 178
d_5 = a_4 + b_4 + c_4 + d_4 = 39 + 39 + 50 + 50 = 178
Therefore, the total number of strings of length 5 that satisfy the given condition is a_5 + b_5 + c_5 + d_5 = 139 + 139 + 178 + 178 = 634.
So, there are 634 strings of length 5 consisting solely of the digits 1, 2, 3, and 4, where any 1 and 2 are never neighbors.
```

`Now, summarize the answer above in one sentence, without any intermediate steps or explanations.`

`There are 634 strings of length 5 consisting solely of the digits 1, 2, 3, and 4, where any 1 and 2 are never neighbors.`