How many permutations of the letters ABCDEFGH contain the string HAD?
The correct answer and explanation is :
To find how many permutations of the letters ABCDEFGH contain the string “HAD,” we can treat the string “HAD” as a single block. By doing this, we reduce the problem to finding the number of permutations of fewer items.
Step 1: Treat “HAD” as a block
Since “HAD” must appear together in the permutation, we treat it as a single “letter” or block. So instead of having 8 individual letters (A, B, C, D, E, F, G, H), we now have:
- The “HAD” block
- The remaining individual letters B, C, E, F, G.
This gives us a total of 6 “objects” to permute: {HAD, B, C, E, F, G}.
Step 2: Find the number of permutations of these 6 objects
The number of ways to arrange 6 objects is simply 6 factorial, denoted as 6!6!. Mathematically: 6!=6×5×4×3×2×1=720.6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720.
Step 3: Accounting for the arrangement of letters within “HAD”
Within the block “HAD,” the order of the letters is fixed — “H” must come first, followed by “A,” and then “D.” Therefore, there are no additional permutations inside the block.
Step 4: Conclusion
Since the block “HAD” has to appear together in every arrangement and no additional rearrangement of the letters within “HAD” is allowed, the total number of permutations where “HAD” appears as a string is simply 6!=7206! = 720.
Thus, the number of permutations of the letters ABCDEFGH that contain the string “HAD” is 720.
Key Points:
- The string “HAD” is treated as a single block.
- The number of ways to arrange the 6 objects is 6!6!.
- The letters inside the “HAD” block are fixed and do not contribute additional permutations.