Find the number of ways of selecting 4 letters from the word EXAMINATION.

Options

A. 136
B. 126
C. 252
D. 525

Answer

To solve the problem of selecting 4 letters from the word "EXAMINATION", we need to consider the frequency of each letter in the word and the different cases that can arise based on these frequencies.

### Step-by-Step Solution:

1. **Identify the letters and their frequencies**:
The word "EXAMINATION" contains the following letters:
- E: 1
- X: 1
- A: 2
- M: 1
- I: 2
- N: 2
- T: 1
- O: 1

Total distinct letters = 8 (E, X, A, M, I, N, T, O).

2. **Determine the cases for selecting 4 letters**:
We can have different scenarios based on the frequency of the letters:
- Case 1: 4 different letters
- Case 2: 2 letters are the same and 2 letters are different
- Case 3: 2 letters are the same and the other 2 letters are also the same (not possible here since we don't have 4 of the same letter)
- Case 4: 2 letters are the same and 2 letters are different (this case can occur with letters that appear twice).

3. **Calculate the number of ways for each case**:

**Case 1: 4 different letters**
We can choose 4 different letters from the 8 distinct letters. The number of ways to choose 4 letters from 8 is given by the combination formula:
\[ \text{Number of ways} = \binom{8}{4} = \frac{8!}{4!(8-4)!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70 \]

**Case 2: 2 letters are the same and 2 letters are different**
Here, we can choose one letter that appears twice (A, I, or N) and then choose 2 different letters from the remaining letters.

- Choose 1 letter from {A, I, N} (3 choices).
- Choose 2 different letters from the remaining 7 letters (since one letter is already chosen).
\[ \text{Number of ways} = 3 \times \binom{7}{2} = 3 \times \frac{7 \times 6}{2 \times 1} = 3 \times 21 = 63 \]

4. **Total number of ways**:
Now, we add the number of ways from both cases:
\[ \text{Total} = 70 + 63 = 133 \]

### Final Answer:
The total number of ways to select 4 letters from the word "EXAMINATION" is **133**.