Question

The number of different permutations that can be formed by taking 4 letters at a time from the letters of the word "REPETITION" is:

• A. 1380
• B. 1218
• C. 1398
• D. 1286

Hint:

For permutations with repeated elements, use the formula \( \frac{n!}{p_1! p_2! \cdots p_k!} \) where \( n \) is the total number of elements and \( p_i \) is the frequency of each repeated element.

Answer 1

The Correct Option is C

We are required to find the number of different permutations that can be formed by taking 4 letters at a time from the word "REPETITION". 

Step 1: Identify the frequency of each letter 
The given word "REPETITION" has 10 letters in total. The frequency of each letter is as follows: - R = 1 - E = 2 - P = 1 - T = 2 - I = 2 - O = 1 - N = 1 

Step 2: Case Analysis 
We need to count the number of valid 4-letter arrangements. We'll use combinations to select letters and account for repeated letters. ### Case 1: All 4 letters are distinct - Choose 4 distinct letters out of 7 distinct letters (R, E, P, T, I, O, N). \[ \text{Ways} = \binom{7}{4} \times 4! = 35 \times 24 = 840 \] --- ### Case 2: 2 letters the same, 2 other distinct letters - Choose 1 letter that appears at least twice (E, T, or I) in \( \binom{3}{1} \). - Choose 2 more distinct letters from the remaining 6 distinct letters in \( \binom{6}{2} \). - Arrange these 4 letters: \[ \text{Ways} = \binom{3}{1} \times \binom{6}{2} \times \frac{4!}{2!} = 3 \times 15 \times 12 = 540 \] --- ### Case 3: Two pairs of identical letters - Choose 2 letters that appear at least twice from \( E, T, I \) in \( \binom{3}{2} = 3 \). - Arrange these 4 letters: \[ \text{Ways} = \binom{3}{2} \times \frac{4!}{2! \times 2!} = 3 \times 6 = 18 \] --- ### Case 4: 3 letters the same, 1 distinct letter
- Choose 1 letter that appears at least twice in \( \binom{3}{1} = 3 \).
- Choose 1 distinct letter from the remaining 6 distinct letters in \( \binom{6}{1} = 6 \).
- Arrange the 4 letters: \[ \text{Ways} = \binom{3}{1} \times \binom{6}{1} \times \frac{4!}{3!} = 3 \times 6 \times 4 = 72 \] --- ### Case 5: 4 identical letters - This is not possible because no letter appears 4 times in the word. 

Step 3: Total Number of Permutations 
Adding all the valid cases: \[ 840 + 540 + 18 = 1398 \]