Show that `""^(n)P_(n)=""^(n)P_(n-1)` for all natural numbers n.

Answer

To show that \( nP_n = nP_{n-1} \) for all natural numbers \( n \), we will use the formula for permutations.

### Step-by-step Solution:

1. **Understanding the Permutation Formula**:
The formula for permutations \( nP_r \) is given by:
\[ nP_r = \frac{n!}{(n-r)!} \]

2. **Calculate \( nP_n \)**:
Here, we set \( r = n \):
\[ nP_n = \frac{n!}{(n-n)!} = \frac{n!}{0!} \]
Since \( 0! = 1 \), we have:
\[ nP_n = \frac{n!}{1} = n! \]

3. **Calculate \( nP_{n-1} \)**:
Now, we set \( r = n-1 \):
\[ nP_{n-1} = \frac{n!}{(n-(n-1))!} = \frac{n!}{1!} \]
Since \( 1! = 1 \), we have:
\[ nP_{n-1} = \frac{n!}{1} = n! \]

4. **Comparing Both Results**:
From the calculations above, we have:
\[ nP_n = n! \quad \text{and} \quad nP_{n-1} = n! \]
Therefore, we can conclude that:
\[ nP_n = nP_{n-1} \]

5. **Conclusion**:
Since both sides are equal, we have shown that \( nP_n = nP_{n-1} \) for all natural numbers \( n \).