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!