The value of `sum_(r=1)^(n)(""^(n)P_(r))/(r!)` is

Options

A. `2^(n)`
B. `2^(n)-1`
C. `2^(n)-1`
D. `2^(n)+1`

Answer

To solve the problem of finding the value of the summation \( \sum_{r=1}^{n} \frac{nP_r}{r!} \), we can follow these steps:

### Step 1: Understand the notation
The notation \( nP_r \) represents the number of permutations of \( n \) items taken \( r \) at a time. It is given by the formula:
\[ nP_r = \frac{n!}{(n-r)!} \]

### Step 2: Rewrite the summation
We can rewrite the summation using the definition of \( nP_r \):
\[ \sum_{r=1}^{n} \frac{nP_r}{r!} = \sum_{r=1}^{n} \frac{n!}{(n-r)! \cdot r!} \]

### Step 3: Recognize the combination
Notice that the term \( \frac{n!}{(n-r)! \cdot r!} \) is the definition of combinations, specifically \( nC_r \):
\[ \frac{n!}{(n-r)! \cdot r!} = nC_r \]
Thus, we can rewrite our summation as:
\[ \sum_{r=1}^{n} nC_r \]

### Step 4: Use the binomial theorem
According to the binomial theorem, we know that:
\[ \sum_{r=0}^{n} nC_r = 2^n \]
This represents the sum of all combinations of \( n \) items.

### Step 5: Adjust for the summation limits
Since our summation starts from \( r=1 \), we can express it as:
\[ \sum_{r=1}^{n} nC_r = \sum_{r=0}^{n} nC_r - nC_0 \]
Where \( nC_0 = 1 \). Therefore, we have:
\[ \sum_{r=1}^{n} nC_r = 2^n - 1 \]

### Final Result
Thus, the value of the original summation \( \sum_{r=1}^{n} \frac{nP_r}{r!} \) is:
\[ 2^n - 1 \]