Answer :
Answer:
338,400
Step-by-step explanation:
A hexadecimal number is a base-16 numeral system representation using sixteen distinct symbols, typically consisting of digits 0-9 and letters A-F, used to express values in computing and digital systems.
To determine the number of 5-digit hexadecimal numbers with exactly two letters, we can consider two cases: one where the first digit is a letter and another where the first digit is a non-letter.
[tex]\dotfill[/tex]
Case 1
In this case, the first digit is a letter. So, there are 4 positions available for the second letter. Therefore, we can choose any 1 out of these 4 positions for the second letter. This is represented by:
[tex]\displaystyle \binom{4}{1}[/tex]
There are 6 possibilities for each of the two letters (A-F):
[tex]\displaystyle \binom{4}{1}\times 6^2[/tex]
After choosing the positions for the letters, the remaining positions are implicitly determined for the non-letter digits, so we don't need to separately consider their positions in the calculation. For each of the remaining three non-letter digits (0-9), there are 10 possibilities:
[tex]\displaystyle \binom{4}{1}\times 6^2\times 10^3[/tex]
So, the total number of possibilities for Case 1 is:
[tex]\displaystyle \binom{4}{1} \times 6^2 \times 10^3 = 144000[/tex]
[tex]\dotfill[/tex]
Case 2
In this case, the first digit is not a letter but also not zero. So, there are 4 positions available for the two letters. Therefore, we can choose any 2 out of these 4 positions for the second letter. This is represented by:
[tex]\displaystyle \binom{4}{2}[/tex]
There are 6 possibilities for each of the two letters (A-F):
[tex]\displaystyle \binom{4}{2}\times 6^2[/tex]
After choosing the positions for the letters, the remaining positions are implicitly determined for the non-letter digits, so we don't need to separately consider their positions in the calculation.
As the first digit is not a letter but also not zero, there are 9 possibilities for the first digit (1-9):
[tex]\displaystyle \binom{4}{2}\times 6^2 \times 9[/tex]
For each of the remaining two non-letter digits (0-9), there are 10 possibilities.
[tex]\displaystyle \binom{4}{2}\times 6^2 \times 9\times 10^2[/tex]
So, the total number of possibilities for Case 2 is:
[tex]\displaystyle \binom{4}{2}\times 6^2 \times 9\times 10^2=194400[/tex]
[tex]\dotfill[/tex]
To find the total number of 5-digit hexadecimal numbers that have exactly two letters, add the possibilities for each case together:
[tex]144000 + 194400 = 338400[/tex]
So, the total number of 5-digit hexadecimal numbers with exactly two letters is:
[tex]\LARGE\boxed{\boxed{338400}}[/tex]
[tex]\dotfill[/tex]
Additional Notes
Multiplication is used to find the total number of possibilities when considering multiple independent choices. Exponents are used to represent repeated multiplication or the number of possibilities for each choice.
