Final answer:
There are different numbers of permutations based on the given strings.
Step-by-step explanation:
The number of permutations of the letters ABCDEFGH with no repeated letters is equal to the factorial of the number of letters. In this case, there are 8 letters, so there are 8! = 8*7*6*5*4*3*2*1 = 40320 permutations.
a. To find the number of permutations that contain the string ED, we can treat ED as a single letter. So, we have 7 letters to arrange, which gives us 7! = 7*6*5*4*3*2*1 = 5040 permutations.
b. Similarly, for the string CDE, we treat it as a single letter. So, we have 6 letters to arrange, which gives us 6! = 6*5*4*3*2*1 = 720 permutations.
c. To find the number of permutations that contain the strings BA and FGH, we treat BA and FGH as separate entities. So, we have 6 letters to arrange, which gives us 6! = 720 permutations.
d. To find the number of permutations that contain the strings AB, DE, and GH, we treat AB, DE, and GH as separate entities. So, we have 5 letters to arrange, which gives us 5! = 120 permutations.
e. To find the number of permutations that contain the strings CAB and BED, we treat CAB and BED as separate entities. So, we have 5 letters to arrange, which gives us 5! = 120 permutations.
f. Similarly, for the strings BCA and ABF, we have 5 letters to arrange, which gives us 5! = 120 permutations.