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Let's work through the problem together.

### Part a: How many 4-digit hexadecimals are there in which the first digit is E or F?

In a hexadecimal system, the digits can be one of 16 possible values: `{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F}`. We are looking for 4-digit hexadecimal numbers where the first digit is either E or F.

- The first digit has 2 choices: either E or F.
- The remaining 3 digits can each be any of the 16 hexadecimal digits (0-9 and A-F), so there are 16 choices for each of these digits.

Thus, the total number of 4-digit hexadecimals with the first digit as E or F is:
\[
2 \times 16 \times 16 \times 16 = 2 \times 16^3 = 2 \times 4096 = 8192
\]

### Part b: How many 3-digit hexadecimals start with a letter (A-F) and end with a numeral (0-9)?

For 3-digit hexadecimal numbers:
- The first digit must be one of the letters A, B, C, D, E, or F, so there are 6 choices.
- The second digit can be any of the 16 hexadecimal digits, so 16 choices.
- The third (last) digit must be one of the numerals 0 through 9, so there are 10 choices.

The total number of 3-digit hexadecimals that start with a letter and end with a numeral is:
\[
6 \times 16 \times 10 = 960
\]

### Part c: How many 2-digit hexadecimals start with a letter (A-F) or end with a numeral (0-9) (or both)?

We need to consider the possibilities for 2-digit hexadecimals under different conditions.

1. **Start with a letter (A-F):**
   - The first digit can be one of A, B, C, D, E, or F, so 6 choices.
   - The second digit can be any of the 16 hexadecimal digits, so 16 choices.
   - Total for this case: \(6 \times 16 = 96\).

2. **End with a numeral (0-9):**
   - The first digit can be any of the 16 hexadecimal digits, so 16 choices.
   - The second digit must be one of the numerals 0 through 9, so 10 choices.
   - Total for this case: \(16 \times 10 = 160\).

3. **Start with a letter and end with a numeral (intersection of both conditions):**
   - The first digit must be one of A, B, C, D, E, or F, so 6 choices.
   - The second digit must be one of the numerals 0 through 9, so 10 choices.
   - Total for this case: \(6 \times 10 = 60\).

Using the principle of inclusion-exclusion:
\[
(96 + 160 - 60) = 196
\]

Thus, the total number of 2-digit hexadecimals that start with a letter (A-F), or end with a numeral (0-9), or both, is 196.

### Summary of answers:
a. \(8192\)  
b. \(960\)  
c. \(196\)

Would you like further details on any of these parts? 

Here are some related questions:
1. How would the result change if we were dealing with 5-digit hexadecimal numbers?
2. Can you calculate how many 6-digit hexadecimals start with a number between 0-7 and end with a letter (A-F)?
3. What if the second digit in part (b) also had to be a letter (A-F)?
4. How would this problem be different if we were working in base 8 (octal)?
5. Can you find how many 3-digit hexadecimal numbers have all digits as letters (A-F)?

**Tip**: In questions involving combinations of conditions (like part c), using inclusion-exclusion helps avoid double-counting situations!

We usually write numbers in decimal form (or base 10), meaning numbers are composed using 10 different 'digits' {0, 1, ..., 9}. Sometimes though it is useful to write numbers hexadecimal or base 16. Now there are 16 distinct digits that can be used to form numbers: {0, 1, ..., 9, A, B, C, D, E, F}. So for example, a 3 digit hexadecimal number might be 2B8.

a. How many 4-digit hexadecimals are there in which the first digit is E or F?
b. How many 3-digit hexadecimals start with a letter (A-F) and end with a numeral (0-9)?
c. How many 2-digit hexadecimals start with a letter (A-F) or end with a numeral (0-9) (or both)?

Explore how to count specific hexadecimal numbers based on given conditions. This problem covers key combinatorics concepts such as the inclusion-exclusion principle and the counting principle, suitable for students in Grades 8-10.

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