Let `theta=(a_(1),a_(2),a_(3),...,a_(n))` be a given arrangement of `n` distinct objects `a_(1),a_(2),a_(3),…,a_(n)`. A derangement of `theta` is an arrangment of these `n` objects in which none of the objects occupies its original position. Let `D_(n)` be the number of derangements of the permutations `theta`.
The relation between `D_(n)` and `D_(n-1)` is given by

Options

A. `D_(n)-nD_(n-1)=(-1)^(n)`
B. `D_(n)-(n-1)D_(n-1)=(-1)^(n-1)`
C. `D_(n)-nD_(n-1)=(-1)^(n-1)`
D. `D_(n)-D_(n-1)=(-1)^(n-1)`

Explanation

To find the relationship between \( D_n \) (the number of derangements of \( n \) distinct objects) and \( D_{n-1} \), we can use the principle of inclusion-exclusion and some combinatorial reasoning. Let's derive the formula step by step.

### Step 1: Understanding Derangements
A derangement is a permutation of a set where none of the elements appear in their original position. For \( n \) distinct objects \( a_1, a_2, \ldots, a_n \), we denote the number of derangements as \( D_n \).

### Step 2: Recursive Relation
To derive the relationship between \( D_n \) and \( D_{n-1} \), consider the placement of the \( n \)-th object \( a_n \):
- If \( a_n \) is placed in the \( r \)-th position (where \( r \) can be any position from 1 to \( n \)), then we have two cases:
1. If \( a_n \) is placed in the \( r \)-th position, then the object originally in the \( r \)-th position (let's call it \( a_r \)) cannot go to its original position. It can either go to the \( n \)-th position or any of the other \( n-1 \) positions.

### Step 3: Counting Derangements
- If \( a_n \) goes to the \( r \)-th position, there are \( D_{n-1} \) ways to derange the remaining \( n-1 \) objects (since \( a_r \) cannot go to its original position).
- If \( a_n \) goes to the \( n \)-th position, then there are \( D_{n-2} \) ways to derange the remaining \( n-1 \) objects (since \( a_r \) can go to any of the \( n-1 \) positions except its own).

### Step 4: Formulating the Recursive Relation
Thus, we can express \( D_n \) as:
\[ D_n = (n-1) \cdot D_{n-1} + D_{n-2} \]
This accounts for all possible placements of the \( n \)-th object and the resulting derangements of the remaining objects.

### Step 5: Conclusion
The relationship between \( D_n \) and \( D_{n-1} \) is given by:
\[ D_n = (n-1)D_{n-1} + D_{n-2} \]