Given are six `0'`s, five `1'`s and four `2's` . Consider all possible permutations of all these numbers. [A permutations can have its leading digit `0`].
How many permutations have the first `0` preceding the first `1` ?

Options

A. `"^(15)C_(4)xx^(10)C_(5)`
B. `"^(15)C_(5)xx^(10)C_(4)`
C. `"^(15)C_(6)xx^(10)C_(5)`
D. `"^(15)C_(5)xx^(10)C_(5)`

Explanation

To solve the problem of finding how many permutations have the first `0` preceding the first `1`, we can follow these steps:

### Step-by-Step Solution:

1. **Count the Total Digits**:
We have 6 `0`s, 5 `1`s, and 4 `2`s. Therefore, the total number of digits is:
\[ 6 + 5 + 4 = 15 \]

2. **Fix the First Digit**:
Since we want the first `0` to precede the first `1`, we can fix the first digit as `0`. This means we have already placed one `0`, leaving us with:
- 5 `0`s remaining
- 5 `1`s
- 4 `2`s

3. **Calculate Remaining Positions**:
After placing the first `0`, we have:
\[ 15 - 1 = 14 \text{ positions left} \]

4. **Choose Positions for `2`s**:
We need to choose positions for the `4` `2`s from the `14` remaining positions. The number of ways to choose `4` positions from `14` is given by the combination:
\[ \binom{14}{4} \]

5. **Calculate Remaining Positions for `0`s and `1`s**:
After placing `4` `2`s, we have:
\[ 14 - 4 = 10 \text{ positions left} \]
Now, we need to place the remaining `5` `0`s and `5` `1`s in these `10` positions.

6. **Choose Positions for `0`s**:
The number of ways to choose `5` positions for the `0`s from the `10` remaining positions is:
\[ \binom{10}{5} \]

7. **Calculate Total Permutations**:
The total number of permutations where the first `0` precedes the first `1` can be calculated by multiplying the number of ways to arrange the `2`s and the number of ways to arrange the remaining `0`s:
\[ \text{Total permutations} = \binom{14}{4} \times \binom{10}{5} \]

8. **Final Calculation**:
Now we can compute the values:
\[ \binom{14}{4} = \frac{14!}{4!(14-4)!} = \frac{14 \times 13 \times 12 \times 11}{4 \times 3 \times 2 \times 1} = 1001 \]
\[ \binom{10}{5} = \frac{10!}{5!(10-5)!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \]
Therefore, the total number of permutations is:
\[ 1001 \times 252 = 252252 \]

### Final Answer:
The total number of permutations where the first `0` precedes the first `1` is **252252**.