Number of permutations of `1,2,3,4,5,6,7,8`, and `9` taken all at a time are such that digit `1` appearing somewhere to the left of `2` and digit `3` appearing to the left of `4` and digit `5` somewhere to the left of `6`, is
(e.g. `815723946` would be one such permutation)

Options

A. `9.7!`
B. `8!`
C. `5!4!`
D. `8!4!`

Explanation

To solve the problem of finding the number of permutations of the digits `1, 2, 3, 4, 5, 6, 7, 8, 9` such that `1` appears to the left of `2`, `3` appears to the left of `4`, and `5` appears to the left of `6`, we can follow these steps:

### Step-by-Step Solution:

1. **Total Permutations**: First, we calculate the total number of permutations of the digits `1` to `9`. The total number of permutations of `n` distinct objects is given by `n!`. Here, `n = 9`, so:
\[ \text{Total permutations} = 9! = 362880 \]

2. **Conditions**: We need to account for the conditions that `1` must be to the left of `2`, `3` must be to the left of `4`, and `5` must be to the left of `6`. Each of these conditions effectively halves the number of valid arrangements since for any two digits, they can either be in one order or the reverse.

3. **Calculating Valid Arrangements**:
- For the pair `1` and `2`, there are 2 arrangements: `1, 2` and `2, 1`. Only `1, 2` is valid, so the probability of `1` being to the left of `2` is \( \frac{1}{2} \).
- For the pair `3` and `4`, the same logic applies. The probability of `3` being to the left of `4` is also \( \frac{1}{2} \).
- For the pair `5` and `6`, the probability of `5` being to the left of `6` is again \( \frac{1}{2} \).

4. **Combining Probabilities**: Since these events are independent, we multiply the probabilities:
\[ P(\text{1 left of 2}) \times P(\text{3 left of 4}) \times P(\text{5 left of 6}) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} \]

5. **Calculating Valid Permutations**: Now, we can find the number of valid permutations by multiplying the total permutations by the probability of the conditions being satisfied:
\[ \text{Valid permutations} = 9! \times \frac{1}{8} = \frac{362880}{8} = 45360 \]

### Final Answer:
The number of permutations of `1, 2, 3, 4, 5, 6, 7, 8, 9` such that `1` is to the left of `2`, `3` is to the left of `4`, and `5` is to the left of `6` is **45360**.