There are 10 different objects 1,2,3…10 arranged at random in 10 planes marked, 1,,2,3,….10. The probability that exactly five of these objects occupy places corresponding to their number is

Options

A. `(13)/(3600)`
B. `(11)/(3600)`
C. `(.^(10)_(6)xx44)/((5!))^2`
D. `(.^(10)C_(5)xx44)/((10!))`

Answer

To find the probability that exactly five of the ten different objects occupy places corresponding to their number, we can follow these steps:

### Step 1: Calculate the Total Number of Outcomes
The total number of ways to arrange 10 different objects in 10 different places is given by the factorial of the number of objects, which is:
\[ \text{Total Outcomes} = 10! \]

### Step 2: Choose 5 Objects That Will Be in Their Correct Places
We need to select 5 objects out of the 10 that will occupy their corresponding positions. The number of ways to choose 5 objects from 10 is given by the combination formula:
\[ \text{Ways to choose 5 objects} = \binom{10}{5} \]

### Step 3: Arrange the Remaining 5 Objects
The remaining 5 objects must not occupy their corresponding positions. To find the number of arrangements where none of the remaining objects are in their correct positions, we can use the principle of inclusion-exclusion.

Let \( D(n) \) denote the number of derangements (permutations where no object appears in its original position) of \( n \) objects. The formula for derangements is:
\[ D(n) = n! \sum_{i=0}^{n} \frac{(-1)^i}{i!} \]
For \( n = 5 \):
\[ D(5) = 5! \left( \frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} \right) \]
Calculating this gives:
\[ D(5) = 120 \left( 1 - 1 + 0.5 - \frac{1}{6} + \frac{1}{24} - \frac{1}{120} \right) \]
\[ = 120 \left( 0 + 0.5 - 0.1667 + 0.0417 - 0.0083 \right) \]
\[ = 120 \left( 0.3667 \right) \approx 44 \]

### Step 4: Calculate the Favorable Outcomes
The total number of favorable outcomes where exactly 5 objects are in their correct places is:
\[ \text{Favorable Outcomes} = \binom{10}{5} \times D(5) \]
Substituting the values:
\[ = \binom{10}{5} \times 44 \]

### Step 5: Calculate the Probability
The probability that exactly 5 objects occupy their corresponding places is given by the ratio of favorable outcomes to total outcomes:
\[ P = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{\binom{10}{5} \times 44}{10!} \]

### Step 6: Simplify the Probability
We know that \( 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5! \), and \( \binom{10}{5} = \frac{10!}{5!5!} \). Thus, we can simplify:
\[ P = \frac{\frac{10!}{5!5!} \times 44}{10!} = \frac{44}{5!} = \frac{44}{120} = \frac{11}{30} \]

### Final Answer
The probability that exactly five of these objects occupy places corresponding to their number is:
\[ \frac{11}{30} \]