To solve the problem, we will evaluate each statement one by one.
### Step 1: Evaluate Statement 1
**Statement 1:** The product of r consecutive integers is always divisible by r!.
Let’s denote the r consecutive integers as \( n, n+1, n+2, \ldots, n+r-1 \). The product of these integers can be expressed as:
\[
P = n(n+1)(n+2)\ldots(n+r-1)
\]
We need to show that \( P \) is divisible by \( r! \).
**Proof:**
1. The factorial \( r! \) is defined as \( r! = r \times (r-1) \times (r-2) \times \ldots \times 1 \).
2. Among any \( r \) consecutive integers, there will be at least one multiple of each integer from \( 1 \) to \( r \). This is because the integers cover all residues modulo \( r \).
3. Therefore, the product \( P \) contains all the factors needed to form \( r! \).
Thus, **Statement 1 is true**.
### Step 2: Evaluate Statement 2
**Statement 2:** The number of ways formed by utilizing all the letters of the word ASSIST in which S's are alternate is 12.
The word ASSIST has the letters A, S, S, I, S, T.
1. The total number of letters is 6.
2. The letter S appears 3 times, and we need to arrange them such that no two S's are adjacent.
3. We can place the other letters (A, I, T) first. The arrangement of A, I, T can be done in \( 3! = 6 \) ways.
4. Once A, I, T are placed, we can place the S's in the gaps created. For example, if we place A, I, T, the arrangement looks like this: _ A _ I _ T _. There are 4 gaps (before A, between A and I, between I and T, and after T).
5. We need to choose 3 out of these 4 gaps to place the S's. This can be done in \( \binom{4}{3} = 4 \) ways.
Thus, the total arrangements are:
\[
6 \times 4 = 24
\]
Since we need the S's to be alternate, we can only use half of these arrangements (as we can choose either odd or even positions for S's). Therefore, the number of valid arrangements is \( \frac{24}{2} = 12 \).
Thus, **Statement 2 is true**.
### Step 3: Evaluate Statement 3
**Statement 3:** The number of words which can be formed out of the letters a, b, c, d, e, f taken 3 together each word containing one vowel at least is 16.
The vowels in the set {a, b, c, d, e, f} are a and e.
1. We can have two cases:
- Case 1: 1 vowel and 2 consonants.
- Case 2: 2 vowels and 1 consonant.
**Case 1: 1 vowel and 2 consonants**
- Choose 1 vowel from {a, e}: \( \binom{2}{1} = 2 \).
- Choose 2 consonants from {b, c, d, f} (4 consonants): \( \binom{4}{2} = 6 \).
- Total combinations for this case: \( 2 \times 6 = 12 \).
- Each selection can be arranged in \( 3! = 6 \) ways, so total = \( 12 \times 6 = 72 \).
**Case 2: 2 vowels and 1 consonant**
- Choose 2 vowels from {a, e}: \( \binom{2}{2} = 1 \).
- Choose 1 consonant from {b, c, d, f}: \( \binom{4}{1} = 4 \).
- Total combinations for this case: \( 1 \times 4 = 4 \).
- Each selection can be arranged in \( 3! = 6 \) ways, so total = \( 4 \times 6 = 24 \).
**Total combinations:**
- Case 1 + Case 2 = \( 72 + 24 = 96 \).
Thus, **Statement 3 is false**.
### Conclusion
- **Statement 1:** True
- **Statement 2:** True
- **Statement 3:** False
The correct sequence is **True, True, False**.
