`P=n(n^(2)-1)(n^(2)-4)(n^(2)-9)…(n^(2)-100)` is always divisible by , `(n in I)` (a) `2!3!4!5!6!` (b) `(5!)^(4)` (c) `(10!)^(2)` (d) `10!11!`

Options

A. `2!3!4!5!6!`
B. `(5!)^(4)`
C. `(10!)^(2)`
D. `10!11!`

Explanation

To solve the problem, we need to analyze the expression \( P = n(n^2 - 1)(n^2 - 4)(n^2 - 9) \ldots (n^2 - 100) \).

### Step 1: Rewrite the expression
We can rewrite the expression \( P \) as:
\[ P = n \cdot (n^2 - 1^2) \cdot (n^2 - 2^2) \cdots (n^2 - 10^2) \]
This can be expressed using the difference of squares:
\[ P = n \cdot (n - 1)(n + 1)(n - 2)(n + 2) \cdots (n - 10)(n + 10) \]

### Step 2: Identify the terms
The terms in \( P \) consist of:
- \( n \)
- \( n - 1, n + 1, n - 2, n + 2, \ldots, n - 10, n + 10 \)

This gives us a total of \( 11 \) terms from \( n - 10 \) to \( n + 10 \).

### Step 3: Count the total number of terms
The total number of terms is \( 1 + 10 + 10 = 21 \) terms (1 from \( n \) and 10 from each side).

### Step 4: Determine divisibility by factorials
The product of \( k \) consecutive integers is divisible by \( k! \). Therefore:
- The product of the \( 11 \) terms \( n, n + 1, n + 2, \ldots, n + 10 \) is divisible by \( 11! \).
- The product of the \( 10 \) terms \( n - 1, n - 2, \ldots, n - 10 \) is divisible by \( 10! \).

Thus, we can conclude:
\[ P \text{ is divisible by } 11! \cdot 10! \]

### Step 5: Analyze the options
Now we will analyze the given options:
- (a) \( 2!3!4!5!6! \) - This is not sufficient.
- (b) \( (5!)^4 \) - This is also not sufficient.
- (c) \( (10!)^2 \) - This is not sufficient.
- (d) \( 10!11! \) - This matches our conclusion.

### Conclusion
The correct answer is:
\[ \text{(d) } 10!11! \]