To solve this, we need to calculate the number of ways the student can select 2 books such that both books are of the same type, either both fiction or both non-fiction.

Step 1: Count the ways to select 2 fiction books.

Since there are 6 fiction books, we can choose 2 of them as follows: $\binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15$

Step 2: Count the ways to select 2 non-fiction books.

Since there are 4 non-fiction books, we can choose 2 of them as follows: $\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6$

Step 3: Add the two results.

The total number of ways to select 2 books of the same type (either both fiction or both non-fiction) is: $15 + 6 = 21$

Final Answer:

The student has 21 ways to select 2 books such that both are either fiction or non-fiction.

Would you like more details or have any questions?

Here are 5 related questions to extend your understanding:

1. What if the student could select 3 books instead of 2?
2. How many ways are there if the student must choose exactly one fiction and one non-fiction book?
3. If a third category, science books, is added with 5 books, how would this affect the calculations?
4. How would the solution change if the selection was required to have at least one fiction book?
5. Can you generalize a formula for choosing books from different categories?

Tip: In combinatorics, the "combination" formula $\binom{n}{r}$ is useful when the order of selection doesn’t matter.