a) 512 codes, b) 216 codes, c) 49 codes, d) 336 codes. Without repetition: a) 36 codes, b) 2401 codes, c) 70 codes, d) 35 codes.
Let's break down each scenario:
For 3-letter codes with repetitions allowed:
a) No restrictions:
Each position has 8 choices (A, B, C, D, E, F, G, H).
Total combinations = 8 * 8 * 8 = 512
b) "A" and "C" cannot be used:
There are 6 choices for each position (B, D, E, F, G, H).
Total combinations = 6 * 6 * 6 = 216
c) "D" must be used:
"D" is fixed in one position, leaving 7 choices for the other two positions.
Total combinations = 1 * 7 * 7 = 49
d) The same letter is not repeated three times:
For the first position, there are 8 choices.
For the second position, there are 7 choices (excluding the letter used in the first position).
For the third position, there are 6 choices (excluding the letters used in the first two positions).
Total combinations = 8 * 7 * 6 = 336
For 4-letter codes without repetition:
a) "A" and "C" must be used:
"A" and "C" are fixed in the first two positions, leaving 6 choices for each of the remaining two positions.
Total combinations = 1 * 1 * 6 * 6 = 36
b) "D" cannot be used:
"D" is excluded, so there are 7 choices for each position.
Total combinations = 7 * 7 * 7 * 7 = 2401
c) The order does not matter:
This is a combination problem, and the formula for combinations is n! / (r! * (n-r)!), where n is the total number of options, and r is the number of choices.
For 4-letter codes without repetition, it is 8! / (4! * (8-4)!) = 8! / (4! * 4!) = 70
d) The order does not matter, and "F" must be used:
Fix "F" in one position, then choose 3 letters from the remaining 7 (excluding "F").
Total combinations = 1 * (7 choose 3) = 1 * 35 = 35