To solve the problem of finding how many 5-letter words can be formed using the letters of the word "CALCULUS", we need to consider the frequency of each letter in the word. The letters in "CALCULUS" are:
- C: 1
- A: 1
- L: 2
- U: 2
- S: 1
Now, let's break down the solution step by step.
### Step 1: Identify the total letters and their frequencies
The word "CALCULUS" has 8 letters in total, with the following frequencies:
- C: 1
- A: 1
- L: 2
- U: 2
- S: 1
### Step 2: Determine the possible cases for forming 5-letter words
We can form 5-letter words in the following cases based on the repetition of letters:
1. **All different letters**: Choose 5 different letters from the available letters.
2. **Two letters are the same, and three are different**: Choose one letter that appears twice and three different letters from the remaining.
3. **Two pairs of letters are the same, and one letter is different**: Choose two letters that appear twice and one different letter from the remaining.
### Step 3: Calculate the number of arrangements for each case
#### Case 1: All different letters
We can choose from C, A, L, U, S. Since we have 5 different letters, we can arrange them in:
\[
5! = 120 \text{ ways}
\]
#### Case 2: Two letters are the same, and three are different
We can choose either L or U to be the letter that appears twice.
- If we choose L:
- We need to select 3 from C, A, U, S (4 letters left).
- The number of ways to choose 3 different letters from 4 is \( \binom{4}{3} = 4 \).
- The arrangement of these 5 letters (2 L's and 3 different) is:
\[
\frac{5!}{2!} = \frac{120}{2} = 60
\]
- Total for this case: \( 4 \times 60 = 240 \).
- If we choose U:
- We follow the same logic as above, which also gives us \( 4 \times 60 = 240 \).
Thus, the total for this case is \( 240 + 240 = 480 \).
#### Case 3: Two pairs of letters are the same, and one letter is different
We can choose L and U to be the letters that appear twice. The remaining letters are C, A, S (3 letters left).
- We can choose 1 letter from C, A, S, which gives us \( \binom{3}{1} = 3 \).
- The arrangement of these 5 letters (2 L's, 2 U's, and 1 different) is:
\[
\frac{5!}{2! \times 2!} = \frac{120}{4} = 30
\]
- Total for this case: \( 3 \times 30 = 90 \).
### Step 4: Sum all cases
Now, we sum the total arrangements from all cases:
\[
120 \text{ (Case 1)} + 480 \text{ (Case 2)} + 90 \text{ (Case 3)} = 690
\]
### Step 5: Check divisibility
Now, we need to check if 690 is divisible by 2, 3, and 5:
- **Divisibility by 2**: The last digit is 0, so it is divisible by 2.
- **Divisibility by 3**: The sum of the digits \( 6 + 9 + 0 = 15 \) is divisible by 3.
- **Divisibility by 5**: The last digit is 0, so it is divisible by 5.
Thus, the number of 5-letter words formed with the letters of the word "CALCULUS" is divisible by 2, 3, and 5.
### Final Answer
The number of 5-letter words formed with the letters of the word CALCULUS is divisible by 2, 3, and 5.
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