For each `n in N` the greatest positive integer which divides `(n + 1) (n + 2) (n + 3) ldots (n + r)` is :

Options

A. r
B. r!
C. `(r + 1)!`
D. `n+r`

Answer

To solve the problem of finding the greatest positive integer that divides the product \((n + 1)(n + 2)(n + 3) \ldots (n + r)\) for each \(n \in \mathbb{N}\), we can follow these steps:

### Step 1: Understand the Expression
The expression \((n + 1)(n + 2)(n + 3) \ldots (n + r)\) consists of \(r\) consecutive integers starting from \(n + 1\) to \(n + r\).

### Step 2: Count the Factors
The product of \(r\) consecutive integers will always be divisible by \(r!\) (r factorial). This is because in any set of \(r\) consecutive integers, there will be at least one multiple of each integer from \(1\) to \(r\).

### Step 3: Verify with Examples
Let’s take an example to verify this. If \(n = 1\) and \(r = 4\), we have:
\[ (1 + 1)(1 + 2)(1 + 3)(1 + 4) = 2 \cdot 3 \cdot 4 \cdot 5 \]
Calculating this gives:
\[ 2 \cdot 3 = 6, \quad 6 \cdot 4 = 24, \quad 24 \cdot 5 = 120 \]
Now, \(4! = 24\), and indeed \(120\) is divisible by \(24\).

### Step 4: Generalize the Result
For any \(n \in \mathbb{N}\), the product \((n + 1)(n + 2)(n + 3) \ldots (n + r)\) will always be divisible by \(r!\). Thus, the greatest positive integer that divides this product is \(r!\).

### Conclusion
The greatest positive integer that divides \((n + 1)(n + 2)(n + 3) \ldots (n + r)\) for each \(n \in \mathbb{N}\) is \(r!\).

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