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Okay, so I need to figure out how to solve this problem: Find the number of ways to arrange the letters of the word "BROCCOLIf" such that no two vowels are adjacent. Hmm, let's start by understanding the problem.
First, I should confirm the letters in "BROCCOLIf". Let me spell it out: B, R, O, C, C, O, L, I, f. Wait, is the last letter lowercase 'f'? The original word is written as "BROCCOLIf". So maybe it's a mix of uppercase and lowercase? But in typical arrangement problems, case matters unless specified otherwise. However, sometimes lowercase and uppercase are considered the same. Wait, the problem says "arrange the letters", so I think case matters here. So the letters are B, R, O, C, C, O, L, I, f. Let me count them: B (1), R (1), O (2), C (2), L (1), I (1), f (1). So total letters: 1+1+2+2+1+1+1=9 letters. Wait, let me recount: B, R, O, C, C, O, L, I, f. Yes, that's 9 letters. But there are repeated letters: O appears twice, C appears twice, and the rest are unique.
Now, the vowels in English are A, E, I, O, U. Sometimes Y is considered a vowel, but I don't think so here. So in "BROCCOLIf", the vowels are O, O, I. Wait, O is a vowel, and I is a vowel. So there are three vowels: two Os and one I. The consonants are B, R, C, C, L, f. So consonants: B, R, C, C, L, f. That's six consonants with C repeated twice.
The problem is to arrange all these letters so that no two vowels are adjacent. So vowels can't be next to each other. The standard approach for such problems is to first arrange the consonants and then place the vowels in the gaps between the consonants.
First, I need to calculate the number of ways to arrange the consonants, considering duplicates. Then, calculate the number of ways to place the vowels in the available gaps, again considering duplicates.
Let me break it down step by step.
Step 1: Arrange the consonants. The consonants are B, R, C, C, L, f. There are 6 consonants with C repeated twice. The number of distinct arrangements is 6! divided by 2! (because of the two Cs). So 6! / 2! = 720 / 2 = 360.
Step 2: Once the consonants are arranged, there are slots (gaps) where vowels can be placed. For 6 consonants, there are 6 + 1 = 7 gaps. These are the positions before the first consonant, between each pair of consonants, and after the last consonant.
We need to place 3 vowels (O, O, I) into these 7 gaps such that no two vowels are in the same gap (since that would make them adjacent). Since the vowels include duplicates (two Os and one I), the number of ways to distribute them is the combination of choosing 3 gaps out of 7, multiplied by the number of distinct arrangements of the vowels in those gaps.
First, choose 3 gaps out of 7: C(7,3) = 35.
Then, arrange the vowels in these 3 gaps. Since there are two Os and one I, the number of distinct arrangements is 3! / 2! = 3.
So total ways to place vowels: 35 * 3 = 105.
Then, multiply the number of consonant arrangements by the number of vowel arrangements: 360 * 105.
Let me compute that: 360 * 100 = 36,000; 360 * 5 = 1,800; total 36,000 + 1,800 = 37,800.
Wait, so the total number of arrangements is 37,800?
But let me double-check. Maybe I missed something with the letters. Let me confirm the letters again.
Original word: B, R, O, C, C, O, L, I, f. So yes, 9 letters. Vowels: O, O, I (3 vowels). Consonants: B, R, C, C, L, f (6 consonants, with two Cs). So consonant arrangements: 6! / 2! = 720 / 2 = 360. Vowel slots: 7. Choosing 3 slots: C(7,3) = 35. Arrangements of vowels: 3 letters with two Os, so 3! / 2! = 3. So 35 * 3 = 105. Total arrangements: 360 * 105 = 37,800.
Hmm, that seems right. But let me think again. Is there any other constraint? For example, does the lowercase 'f' affect anything? Since the problem says "arrange the letters", and the letters include both uppercase and lowercase, but in the original word, it's "BROCCOLIf". Wait, the first letter is uppercase B, then R, O, C, C, O, L, I, and the last letter is lowercase f. So the letters are B, R, O, C, C, O, L, I, f. So the case of 'f' is different from the others. But in terms of arranging, each letter is distinct except for the duplicates. Wait, but 'f' is lowercase, while others are uppercase. So are they considered different? For example, is 'F' and 'f' considered different letters? In standard arrangements, case matters unless specified otherwise. So if the problem is case-sensitive, then 'f' is different from 'F', but in the given word, there's only one 'f' and no 'F's. So in the consonants, we have B, R, C, C, L, I, f. Wait, no, wait: the consonants are B, R, C, C, L, f. So 'f' is lowercase, but the others are uppercase. So in terms of distinctness, 'f' is a different letter from 'F', but since there's no 'F', it's unique. So all consonants except the two Cs are unique. So consonants: B, R, C, C, L, f. So yes, two Cs, others are unique. So the calculation for consonants is correct: 6 letters with two Cs, so 6! / 2! = 360.
Vowels: O, O, I. So two Os and one I. So arranging them in the gaps: 3 positions, with duplicates, so 3! / 2! = 3. Correct.
So total arrangements: 360 * 105 = 37,800.
Wait, but let me check if the vowels are O, O, I. Yes, from the word "BROCCOLIf", the vowels are O (positions 3 and 6) and I (position 8). So three vowels. Correct.
Alternatively, maybe I should consider if the letters are case-sensitive. For example, if 'O' and 'o' are different, but in the given word, all vowels are uppercase except maybe the last 'f', which is consonant. So vowels are O, O, I, all uppercase. So they are considered the same as each other? Wait, no. In arrangements, each letter is distinct based on case. But in this case, the vowels are O, O, I. So two Os and one I. If the case matters, but all vowels here are uppercase, so they are treated as identical letters except for their case. Wait, but in the given word, are the Os uppercase? Yes, in "BROCCOLIf", the letters are B, R, O, C, C, O, L, I, f. So O and I are uppercase, f is lowercase. So the vowels are O, O, I. So two Os and one I. So when arranging, the two Os are identical, and the I is different. So when placing the vowels, we have to account for duplicates. So the calculation is correct: 3! / 2! = 3 ways to arrange the vowels in the selected gaps.
Therefore, the total number of arrangements is 360 * 105 = 37,800.
But wait, let me confirm once more. Let's take a smaller example to see if the method works.
Suppose we have a word with consonants "BC" and vowels "A". Then total letters: 3. The number of arrangements where vowels are not adjacent. Since there's only one vowel, it's just the number of permutations: 3! = 6. But using the method: arrange consonants "BC" (2 letters, no duplicates) → 2! = 2. Then place the vowel in 3 gaps (positions: _ B _ C _). Choose 1 gap out of 3: C(3,1)=3. Then arrange the vowel (only 1 way). So total: 2*3=6. Correct.
Another example: word "BANANA". Let's say we want to arrange letters so that no two vowels are adjacent. Vowels are A, A, A, A (four As). Consonants are B, N, N. So consonants: 3 letters with two Ns. Arrangements: 3! / 2! = 3. Gaps: 4. Choose 4 gaps to place 4 vowels, but since we need to place one vowel in each gap. Since all vowels are identical, the number of ways is C(4,4)=1. So total arrangements: 3*1=3. Which is correct because the only way is to place A between each consonant and at the ends: A B A N A N A. But since consonants can be arranged as B, N, N or N, B, N or N, N, B. So three different arrangements. Correct.
So applying the same logic to the original problem seems valid. Therefore, the answer should be 37,800.
But wait, let me check the multiplication again: 360 * 105. 360 * 100 = 36,000; 360 * 5 = 1,800. 36,000 + 1,800 = 37,800. Yes.
Alternatively, 105 * 360: 100360=36,000; 5360=1,800. Same result.
Therefore, I think the correct answer is 37,800.
To determine the number of ways to arrange the letters of "BROCCOLIf" such that no two vowels are adjacent, follow these steps:
Identify Letters and Vowels:
Total letters: B, R, O, C, C, O, L, I, f (9 letters).
Vowels: O, O, I (3 vowels, with 2 Os and 1 I).
Consonants: B, R, C, C, L, f (6 consonants, with 2 Cs).
Arrange the Consonants:
Calculate permutations of consonants accounting for duplicates:
[
\frac{6!}{2!} = \frac{720}{2} = 360
]
Determine Vowel Placement:
With 6 consonants, there are (6 + 1 = 7) gaps.
Choose 3 gaps out of 7 to place vowels:
[
\binom{7}{3} = 35
]
Arrange the vowels (O, O, I) in the selected gaps, considering duplicates:
[
\frac{3!}{2!} = 3
]
Total vowel placements:
[
35 \times 3 = 105
]
Combine Results:
Multiply consonant arrangements by vowel placements:
[
360 \times 105 = 37,!800
]
Answer: The number of valid arrangements is (\boxed{37800}).
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发表于 2025-11-3 21:26:27