How many permutations of the letters ABCDEFGH contain

How many permutations of the letters ABCDEFGH contain

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?

The correct answer and explanation is :

We are asked to find the number of permutations of the letters ABCDEFGH that contain certain strings. Let’s tackle each part step by step.

General Concept:

A permutation is an arrangement of objects, and when we are given certain constraints, such as specific strings that must appear together, we treat those strings as “blocks” or “super letters.” The number of permutations will depend on how many such “blocks” we have and how many remaining individual letters there are to arrange.

a) The string ED:

If “ED” must appear together, treat “ED” as a single block or “letter.” Therefore, instead of permuting 8 individual letters, we are permuting 7 units: A, B, C, D, E, F, G, and the block ED.

• The number of ways to arrange these 7 units is ( 7! ).
• Within the block “ED”, there is only one way to arrange “E” and “D”, since the order is fixed.
• Thus, the total number of permutations is ( 7! = 5040 ).

b) The string CDE:

Similarly, if “CDE” must appear together, treat “CDE” as a single block. Now, we are permuting 6 units: A, B, F, G, H, and the block CDE.

• The number of ways to arrange these 6 units is ( 6! ).
• Within the block “CDE”, there is only one way to arrange “C”, “D”, and “E”.
• Thus, the total number of permutations is ( 6! = 720 ).

c) The strings BA and FGH:

Now, we have two blocks: “BA” and “FGH”. Treat these as single units, so we are permuting 6 units: C, D, E, and the blocks BA and FGH.

• The number of ways to arrange these 6 units is ( 6! ).
• The “BA” block has only one arrangement, and similarly, the “FGH” block has only one arrangement.
• Thus, the total number of permutations is ( 6! = 720 ).

d) The strings AB, DE, and GH:

Here, we have three blocks: “AB”, “DE”, and “GH”. Treat these as single units, so we are permuting 5 units: C, F, and the three blocks.

• The number of ways to arrange these 5 units is ( 5! ).
• The blocks “AB”, “DE”, and “GH” each have only one arrangement.
• Thus, the total number of permutations is ( 5! = 120 ).

e) The strings CAB and BED:

Now, we have two blocks: “CAB” and “BED”. Treat these as single units, so we are permuting 6 units: F, G, H, and the two blocks.

• The number of ways to arrange these 6 units is ( 6! ).
• The “CAB” block and the “BED” block each have only one arrangement.
• Thus, the total number of permutations is ( 6! = 720 ).

f) The strings BCA and ABF:

We now have two blocks: “BCA” and “ABF”. Treat these as single units, so we are permuting 6 units: D, E, G, H, and the two blocks.

• The number of ways to arrange these 6 units is ( 6! ).
• The “BCA” block and the “ABF” block each have only one arrangement.
• Thus, the total number of permutations is ( 6! = 720 ).

Summary of Results:

• a) 7! = 5040
• b) 6! = 720
• c) 6! = 720
• d) 5! = 120
• e) 6! = 720
• f) 6! = 720

These results are derived from the principle of treating the specified strings as “blocks” and then counting the permutations of the resulting units. The factorials account for the number of ways to arrange these blocks and the remaining individual letters.