Answer
a) 5040
b) 720
c) 120
d) 120
e) 24
f) 0
Work Step by Step
a) We treat the string ED as single letter, thus we have then the possible letters A, B, C, ED, F, G, H. So, The problem is to count permutations of seven items ~ the letters A, B, C, ED, F, G and H. Therefore, the answer is
$$^7P_7= 7! = 5040.$$
b) We treat the string CDE as single letter, thus we have then the possible letters A, B, CDE, F, G, H. So, The problem is to count permutations of six items ~ the letters A, B, CDE, F, G and H Therefore, the answer is
$$^6P_6= 6! = 720.$$
c) We treat the string BA as single letter and FGH as another string, thus we have then the possible letters BA, C, D, E and FGH. So, The problem is to count permutations of five items ~ the letters BA, C, D, E and FGH.. Therefore, the answer is
$$^5P_5= 5! = 120.$$
d) We treat the string AB as single letter, DE as one string and GH as another string, thus we have then the possible letters AB, C, DE, F, GH. So, The problem is to count permutations of five items ~ the letters AB, C, DE, F and GH. Therefore, the answer is
$$^5P_5= 5! = 120.$$
e) If both CAB and BED are substrings, then CABED has to be a substring. So we are really just permuting four items: CABED, F, G and H. Therefore the answer is
$$^4P_4= 4! = 24.$$
f) A string cannot contain both BCA and ABF. Since the string can contain B only once and B cannot be followed directly by both C and F. Thus, there is no string that contain both the string BCA and ABF.
