What is `""^(n-1)P_(r )+ ""^(n-1)P_(r-1)` ?

Answer

To solve the expression \( {}^{n-1}P_r + {}^{n-1}P_{r-1} \), we will use the properties of permutations.

### Step-by-Step Solution:

1. **Understand the Permutation Notation**:
The notation \( {}^{n}P_{r} \) represents the number of ways to choose \( r \) objects from \( n \) distinct objects and arrange them. It is given by the formula:
\[ {}^{n}P_{r} = \frac{n!}{(n-r)!} \]

2. **Write the Given Expression**:
We have the expression:
\[ {}^{n-1}P_r + {}^{n-1}P_{r-1} \]

3. **Substitute the Permutation Formula**:
Now, we can substitute the formula for permutations:
\[ {}^{n-1}P_r = \frac{(n-1)!}{(n-1-r)!} \]
\[ {}^{n-1}P_{r-1} = \frac{(n-1)!}{(n-1-(r-1))!} = \frac{(n-1)!}{(n-r)!} \]

4. **Combine the Two Terms**:
Now we can write the expression as:
\[ {}^{n-1}P_r + {}^{n-1}P_{r-1} = \frac{(n-1)!}{(n-1-r)!} + \frac{(n-1)!}{(n-r)!} \]

5. **Factor Out Common Terms**:
Notice that \( (n-1)! \) is common in both terms:
\[ = (n-1)! \left( \frac{1}{(n-1-r)!} + \frac{1}{(n-r)!} \right) \]

6. **Simplify the Expression Inside the Parentheses**:
To combine the fractions, we need a common denominator:
\[ = (n-1)! \left( \frac{(n-r) + 1}{(n-1-r)!(n-r)} \right) \]
This simplifies to:
\[ = (n-1)! \cdot \frac{n-r}{(n-1-r)!(n-r)} = \frac{n!}{(n-r)!} \]

7. **Final Result**:
Therefore, we have:
\[ {}^{n-1}P_r + {}^{n-1}P_{r-1} = {}^{n}P_r \]

### Conclusion:
Thus, the final answer is:
\[ {}^{n-1}P_r + {}^{n-1}P_{r-1} = {}^{n}P_r \]