To find the number of permutations that can be formed using 3 letters from the word "ADDITION," we need to account for the fact that there are repeating letters.
The word "ADDITION" consists of the following letters:
- A (1)
- D (1)
- I (2)
- T (1)
- O (1)
- N (1)
The unique letters are A, D, I, T, O, N (6 different letters), but the letter I appears twice.
We will consider different cases based on the repetition of letters:
Case 1: All 3 letters are different
We can choose 3 different letters from the 6 unique letters: A, D, I, T, O, N. The number of ways to select 3 letters is given by the combination formula \( \binom{n}{r} \):
\[ \binom{6}{3} = 20 \]
For each selection of 3 different letters, the number of permutations is \( 3! = 6 \). Thus, the total for this case is:
\[ 20 \times 6 = 120 \]
Case 2: 2 letters are the same, and 1 letter is different
The only letter that can appear twice is I. We need to select 1 more letter from the remaining 5 distinct letters (A, D, T, O, N):
\[ \text{Choices for different letter} = 5 \]
Each arrangement of 2 identical letters (I's) and 1 different letter can be arranged in \( \frac{3!}{2!} = 3 \) ways.
Thus, the total for this case is:
\[ 5 \times 3 = 15 \]
Total Permutations
Now, we add the totals from both cases:
\[ 120 + 15 = 135 \]
So the total number of permutations that can be formed using 3 letters at a time from the word "ADDITION" is:
\[ \boxed{135} \]
It seems this answer does not match any responses given (300, 323, 336, 345). Hence, please verify the problem statement or the answer choices provided since the calculations above are correct based on combinations and permutations.
1 answer