To find the number of permutations that can be formed using 3 letters at a time from the letters in the word "ADDITION", we first analyze the composition of the letters.
The word "ADDITION" has the following letter frequencies:
- A: 1
- D: 2
- I: 1
- T: 1
- O: 1
- N: 1
This gives us a total of 8 letters, including the repetition of the letter D.
We will consider different scenarios based on how many of the repeated letter (D) we can use in our combinations of 3 letters:
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All different letters: Here, we can choose 3 different letters from {A, D, I, T, O, N}. The distinct letters are A, D, I, T, O, N which gives us 6 distinct letters.
- Number of ways to choose 3 out of 6: \( \binom{6}{3} = 20 \)
- Each selection of 3 letters can be arranged in \( 3! = 6 \) ways.
- Total for this case: \( 20 \times 6 = 120 \)
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One D (and 2 different letters): We select 1 D and 2 letters from the other distinct letters {A, I, T, O, N} (5 letters).
- Number of ways to choose 2 out of 5: \( \binom{5}{2} = 10 \)
- Each selection can be arranged in \( 3! = 6 \) ways.
- Total for this case: \( 10 \times 6 = 60 \)
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Two Ds (and 1 different letter): We select 2 Ds and 1 letter from the remaining distinct letters {A, I, T, O, N} (5 letters).
- Number of ways to choose 1 out of 5: \( \binom{5}{1} = 5 \)
- Each selection can be arranged in \( 3! / 2! = 3 \) ways (since we have 2 identical letters).
- Total for this case: \( 5 \times 3 = 15 \)
Now we add all the cases together to find the total number of permutations:
\[ 120 + 60 + 15 = 195 \]
Thus, the total number of permutations that can be formed using 3 letters at a time from the letters in "ADDITION" is \( \boxed{195} \).
If this choice is not in the provided options, please double-check or clarify the question for any additional constraints or interpretations.
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