A book contains 1000 pages numbered consecutively. The probability that the sum of the digits of the number of a page is `9, is a/b,` then `a xx b` =

Answer

To solve the problem, we need to find the probability that the sum of the digits of the page number is 9 for a book containing 1000 pages (numbered from 1 to 1000).

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need to find how many page numbers from 1 to 1000 have digits that sum to 9. The page numbers can be represented as three digits (XYZ), where X, Y, and Z are the digits of the page number.

2. **Setting Up the Equation**:
We need to find the number of non-negative integer solutions to the equation:
\[ X + Y + Z = 9 \]
where \(X\), \(Y\), and \(Z\) represent the digits of the page number.

3. **Using the Stars and Bars Method**:
The "stars and bars" theorem helps us find the number of ways to distribute \(n\) indistinguishable objects (stars) into \(k\) distinguishable boxes (variables). In our case, we have 9 stars (the total sum) and 2 bars (to separate the three variables).

The formula for the number of ways to arrange \(n\) stars and \(k-1\) bars is given by:
\[ \binom{n+k-1}{k-1} \]
Here, \(n = 9\) and \(k = 3\) (for \(X\), \(Y\), and \(Z\)). Thus, we need to calculate:
\[ \binom{9 + 3 - 1}{3 - 1} = \binom{11}{2} \]

4. **Calculating the Binomial Coefficient**:
\[ \binom{11}{2} = \frac{11 \times 10}{2 \times 1} = 55 \]
So, there are 55 page numbers where the sum of the digits equals 9.

5. **Calculating the Probability**:
The total number of pages is 1000. Therefore, the probability \(P\) that a randomly selected page number has digits that sum to 9 is:
\[ P = \frac{55}{1000} = \frac{11}{200} \]

6. **Identifying \(a\) and \(b\)**:
Here, \(a = 11\) and \(b = 200\). We need to find the product \(a \times b\):
\[ a \times b = 11 \times 200 = 2200 \]

### Final Answer:
The required answer is \(2200\).