To solve this problem, we can use the principle of inclusion-exclusion. Let's denote the group of students:
- Total students (T) = 25
- Students with cats (C) = 16
- Students with dogs (D) = 15
First, we need to find out how many students have either cats, dogs, or both. The number of students having at least one pet can be expressed using the formula:
\[ |C \cup D| = |C| + |D| - |C \cap D| \]
where \(|C \cap D|\) is the number of students who have both cats and dogs.
However, we don't know \(|C \cap D|\). To find the maximum number of students without any pets, we can minimize \(|C \cap D|\).
The maximum value for \(|C \cap D|\) would be the smallest of the two numbers of pets, which in this case is 15 (the total number of students with dogs). Therefore, the largest possibility for students owning both pets is 15.
If we assume 15 students have both cats and dogs, we can then determine how many have at least one type by substituting:
\[ |C \cup D| = 16 + 15 - 15 = 16 \]
This means that the number of students with at least one pet is 16.
Now we find out how many do not have any pets:
\[ \text{Students without any pets} = T - |C \cup D| = 25 - 16 = 9 \]
So, 9 students do not have any pets.