The value of
`(1.2.3…..9).(11.12.13…19).(21.22.23….29).(31.32.33……39).(41.42.43….49).(51.52.53…..59)……(91.92….99)`

Options

A. `(100!)/(36288 xx 10^(12))`
B. `(99!)/(388 xx 10^(11))`
C. `(99!)/(36288 xx 10^(10))`
D. can't be determined

Answer

To solve the expression `(1.2.3…..9).(11.12.13…19).(21.22.23….29).(31.32.33……39).(41.42.43….49).(51.52.53…..59)……(91.92….99)`, we need to analyze the pattern and fill in the missing numbers systematically.

### Step-by-Step Solution:

1. **Identify the Pattern**:
The expression consists of products of sequences of numbers. Each sequence is missing the numbers that are multiples of 10 (i.e., 10, 20, 30, ..., 90).

2. **List the Complete Sequence**:
The complete sequence from 1 to 99 is:
`1, 2, 3, ..., 9, 10, 11, ..., 19, 20, ..., 29, 30, ..., 39, 40, ..., 49, 50, ..., 59, 60, ..., 69, 70, ..., 79, 80, ..., 89, 90, ..., 99`.

3. **Identify Missing Numbers**:
The missing numbers are: `10, 20, 30, 40, 50, 60, 70, 80, 90`.

4. **Multiply and Divide by Missing Numbers**:
To account for the missing numbers, we can multiply and divide the expression by these numbers:
\[ \text{Expression} = \frac{(1 \cdot 2 \cdot 3 \cdots 9 \cdot 10)(11 \cdots 19 \cdots 20)(21 \cdots 29 \cdots 30)(31 \cdots 39 \cdots 40)(41 \cdots 49 \cdots 50)(51 \cdots 59 \cdots 60)(61 \cdots 69 \cdots 70)(71 \cdots 79 \cdots 80)(81 \cdots 89 \cdots 90)(91 \cdots 99)}{10 \cdot 20 \cdot 30 \cdots 90} \]

5. **Simplify the Expression**:
The numerator can be expressed as \(100!\) (the factorial of 100), and the denominator can be expressed as \(10 \cdot 20 \cdot 30 \cdots 90 = 10^9 \cdot (1 \cdot 2 \cdot 3 \cdots 9) = 10^9 \cdot 9!\).

6. **Final Expression**:
Thus, we can rewrite the expression as:
\[ \frac{100!}{10^9 \cdot 9!} \]

7. **Calculate the Final Value**:
We know that \(9! = 362880\), so:
\[ \frac{100!}{10^9 \cdot 362880} \]

8. **Final Result**:
This gives us the final value of the expression.