The first eight letters of the English alphabet are arranged at random. What is the probability that the letters b, c, d and e always appear together as one group (in any order) in the arrangement?

Introduction / Context:
This question applies permutations and probability to letters of the alphabet. We consider random arrangements of the first eight letters A, B, C, D, E, F, G and H and require that the four letters b, c, d and e appear together as a single block, in any internal order, within the full arrangement.

Given Data / Assumptions:

• Letters used: A, B, C, D, E, F, G, H (8 distinct letters).
• All orderings of these eight letters are equally likely.
• The block condition requires that B, C, D and E occupy four consecutive positions, but within that block their order may vary.
• We seek the probability of this event.

Concept / Approach:
We model the block of four letters as a single combined object. Then, together with the remaining letters, we count how many permutations satisfy the block condition. Inside the block, the four letters can be permuted in 4! ways. The probability will be the ratio of favourable permutations to the total permutations, which is 8! for eight distinct letters.

Step-by-Step Solution:
Total permutations of 8 distinct letters = 8!.Treat the letters B, C, D and E as one block. Now the objects are: [BCDE], A, F, G, H. That is 5 objects to arrange.Permutations of these 5 distinct objects = 5!.Inside the block, the letters B, C, D and E can be arranged in 4! ways.Therefore, the number of favourable permutations = 5! * 4!.Probability = (5! * 4!) / 8!.Compute: 5! = 120, 4! = 24, 8! = 40320, so favourable = 120 * 24 = 2880.Probability = 2880 / 40320 = 1 / 14 after simplification.

Verification / Alternative check:
We can simplify algebraically. Probability = (5! * 4!) / 8! = (120 * 24) / 40320. Divide numerator and denominator by 240: 2880 / 40320 = 12 / 168 = 1 / 14. Performing this purely symbolically confirms the same reduced fraction, giving confidence in the answer.

Why Other Options Are Wrong:
The option 1/7 is exactly double the correct probability and typically arises from forgetting that the block has four letters and miscounting the number of positions. Options 8! and 7! are raw factorial counts, not probabilities, and do not lie between 0 and 1, so they cannot represent a valid probability. Only 1/14 matches the correctly reduced ratio of favourable permutations to all permutations.

Common Pitfalls:
Students often confuse the probability with the count of favourable arrangements. They might stop after computing 5! * 4! and choose 2880 as an answer, or they might divide incorrectly by 7! instead of 8!. Another mistake is assuming that the letters must appear in the specific order BCDE, rather than any permutation of those four letters, which would miss the 4! factor in the count. Always separate the counting of block positions from the internal arrangements of the block.

Final Answer:
The required probability is 1/14.