you are to set up a code of consisting of three letters (A to Z) and five numbers (0 to 9).


a) how many codes are possible if the letters and numbers can be repeated?
b) how many codes are possible if each letter and digit can be only used once?
c) how many codes are possible if each code begins with an A and if the digits cannot be repeated?

B) it would be 26P4 (the number of permutations for the letters) x 9P3 (number of permutations for the digits)

You times because for each letter permutation the can be 9P3 digit permutations.

a) To calculate the number of possible codes when both letters and numbers can be repeated, we need to find the total number of possibilities for each position (3 letters and 5 numbers) and multiply them together.

For the letters, we have 26 options (A to Z). Since repetition is allowed, the number of possibilities for each position is also 26.

For the numbers, we have 10 options (0 to 9). Again, repetition is allowed, so we have 10 possibilities for each position.

To find the total number of possible codes in this case, we multiply the possibilities for each position:

Total possibilities = 26^3 * 10^5 = 175,760,000 codes

Therefore, there are 175,760,000 possible codes if both letters and numbers can be repeated.

b) If each letter and digit can only be used once, the calculation will be slightly different.

For the letters, we still have 26 options for the first letter, but for the second letter, we have 25 options left (since one letter has already been used). Similarly, for the third letter, we will have 24 options left.

For the numbers, the first number will have 10 options. For the second number, we have 9 options left, and so on.

To find the total number of possible codes in this case, we need to multiply the possibilities for each position:

Total possibilities = 26 * 25 * 24 * 10 * 9 * 8 * 7 * 6 = 112,320,000 codes

Therefore, there are 112,320,000 possible codes if each letter and digit can only be used once.

c) If each code must begin with the letter A and digits cannot be repeated, the calculations will be similar to part b), but with some adjustments.

For the first letter, we have only one option, which is A.

For the second letter, since the first letter is fixed at A, we still have 25 options left.

For the third letter, we have 24 options left.

For the numbers, the first number will have 10 options.

For the second number, we have 9 options left (since repetition is not allowed), and so on.

To find the total number of possible codes in this scenario, we need to multiply the possibilities for each position:

Total possibilities = 1 * 25 * 24 * 10 * 9 * 8 * 7 * 6 = 2,646,000 codes

Therefore, there are 2,646,000 possible codes if each code begins with an A and the digits cannot be repeated.