A mapping is selected at random from set `A={1, 2, ...., 10}` into itself. The probability that mapping selected is an injective is

Options

A. `(10)/(10^(9))`
B. `(9!)/(10^(9))`
C. `(9)/(10!)`
D. None of these

Answer

To solve the problem of finding the probability that a randomly selected mapping from the set \( A = \{1, 2, \ldots, 10\} \) into itself is injective, we can follow these steps:

### Step 1: Understand the Concept of Injective Mapping
An injective mapping (or one-to-one mapping) means that each element in the domain maps to a unique element in the codomain. In other words, no two elements in the domain can map to the same element in the codomain.

### Step 2: Determine the Total Number of Mappings
The total number of mappings from set \( A \) to itself can be calculated as follows:
- Each of the 10 elements in set \( A \) can be mapped to any of the 10 elements in set \( A \).
- Therefore, the total number of mappings is \( 10^{10} \).

### Step 3: Calculate the Number of Injective Mappings
To find the number of injective mappings:
- For the first element (1), there are 10 choices (it can map to any of the 10 elements).
- For the second element (2), there are 9 choices (it cannot map to the element that 1 mapped to).
- For the third element (3), there are 8 choices, and so on.
- This continues until the last element (10), which has only 1 choice left.

Thus, the number of injective mappings can be expressed as:
\[ 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 10! \]

### Step 4: Calculate the Probability
The probability \( P \) that a randomly selected mapping is injective can be calculated using the formula:
\[ P = \frac{\text{Number of Injective Mappings}}{\text{Total Number of Mappings}} = \frac{10!}{10^{10}} \]

### Step 5: Simplify the Probability
We can simplify the expression:
\[ P = \frac{10!}{10^{10}} = \frac{10 \times 9!}{10^{10}} = \frac{9!}{10^9} \]

### Final Result
Thus, the probability that a randomly selected mapping from set \( A \) into itself is injective is:
\[ P = \frac{9!}{10^9} \]