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Okay, so I need to figure out how to solve this problem: "EVERY". Wait, the user just wrote "EVERY" and then said "Please reason step by step..." Hmm, maybe they meant to ask a question involving the word "EVERY" but it got cut off? Let me check again. The initial message is "EVERY" followed by the request for a step-by-step explanation. Maybe the problem is related to permutations, combinations, or something with the word "EVERY" itself?
Alternatively, could it be a typo or an incomplete question? If I consider "EVERY" as a word, maybe it's about arranging the letters or finding the number of anagrams? Let me think. The word "EVERY" has 5 letters. Let's see: E, V, E, R, Y. Wait, there are repeating letters here. The letter E appears twice, and the rest are unique. So if the task is to find the number of distinct arrangements of the letters in "EVERY", that would be a permutation problem with repeated elements.
The formula for permutations of multiset is n! divided by the product of the factorials of the counts of each repeated element. So in this case, n is 5 (total letters), and the repeated element is E, which appears twice. So the number of distinct anagrams would be 5! / 2! = 120 / 2 = 60. Is that right? Let me verify. For example, if all letters were unique, it would be 5! = 120. But since there are two E's, each unique arrangement is counted twice (once for each E), so we divide by 2! to account for the duplicates. Yeah, that makes sense. So the answer would be 60.
But wait, maybe the problem is different. Maybe it's a math problem that uses the word "EVERY" in context. Like, "Every number in the set..." but the user didn't finish the question. Hmm. Without more context, it's hard to tell. But given that the user only wrote "EVERY", and given that they might have intended a permutation question, I'll go with that. Let me just make sure there's no other interpretation.
Alternatively, could it be a question about binary operations or something else? For example, "Every element in a group..." but again, without more context, it's unclear. Since the user hasn't provided a specific problem, but just the word "EVERY", and given that permutations with repeated letters is a common question, I think that's the most likely scenario. So I'll proceed with calculating the number of distinct anagrams of the word "EVERY".
To determine the number of distinct anagrams of the word "EVERY," we use the permutation formula for multiset:
Identify the letters and their counts:
Letters: E, V, E, R, Y
Total letters (n): 5
Repeated letters: E appears 2 times; V, R, Y are unique.
Apply the permutation formula:
[
\frac{n!}{\text{(product of factorials of repeated elements)}}
]
Substituting the values:
[
\frac{5!}{2!} = \frac{120}{2} = 60
]
Answer: There are 60 distinct anagrams of the word "EVERY."
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发表于 2025-11-10 02:49:12