Question

How many permutations of the letters ABCDEFGH contain (no letters are repeated) (12 pts)? a. The string ED? b. The string CDE? c. The strings BA and FGH? d. The strings AB, DE, and GH? e. The strings CAB and BED? f. The strings BCA and ABF?

Answer 1

The question involves finding permutations of the letters ABCDEFGH that include specific strings. Each string sequence is considered a single unit, and permutations are then calculated for the remaining letters and these units.

Step-by-step explanation:

The question is asking for the number of permutations of the string ABCDEFGH that contain certain specified sequences like ED, CDE, BA and FGH, AB, DE, and GH, CAB and BED, as well as BCA and ABF. For each specific case, we treat the specified sequences as a single unit and permute amongst the remaining letters and the units.

Calculations:

a. The string ED: Treating 'ED' as a single unit, we have 7 units total (ED and A, B, C, F, G, H). Permutations of 7 units are 7!.

b. The string CDE: Treating 'CDE' as a single unit, we have 6 units total. Permutations are 6!.

c. The strings BA and FGH: Treating 'BA' and 'FGH' as single units, we have 5 units total. Permutations are 5!.

d. The strings AB, DE, and GH: Treating 'AB', 'DE', and 'GH' each as a single unit, we have 5 units total. Permutations are 5!.

e. The strings CAB and BED: Treating 'CAB' and 'BED' as single units, we have 4 units total. Permutations are 4!.

f. The strings BCA and ABF: Treating 'BCA' and 'ABF' as single units, we have 4 units total. Permutations are 4!.

Answer 2

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.