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, we need to find the value of the summation:

\[ \sum_{r=1}^{n} \frac{nP_r}{r!} \]

### Step 1: Understanding the term \(\frac{nP_r}{r!}\)

The term \(nP_r\) represents the number of permutations of \(r\) distinct objects chosen from \(n\) distinct objects. It can be expressed as:

\[ nP_r = \frac{n!}{(n-r)!} \]

Thus, we can rewrite \(\frac{nP_r}{r!}\) as:

\[ \frac{nP_r}{r!} = \frac{n!}{(n-r)! \cdot r!} \]

### Step 2: Recognizing the binomial coefficient

The expression \(\frac{n!}{(n-r)! \cdot r!}\) is the definition of the binomial coefficient \(nC_r\). Therefore, we can rewrite our summation as:

\[ \sum_{r=1}^{n} \frac{nP_r}{r!} = \sum_{r=1}^{n} nC_r \]

### Step 3: Summation of binomial coefficients

We know from the binomial theorem that:

\[ \sum_{r=0}^{n} nC_r = 2^n \]

This summation includes \(nC_0\) (which is 1). However, our summation starts from \(r=1\). To adjust for this, we can express our summation as:

\[ \sum_{r=1}^{n} nC_r = \sum_{r=0}^{n} nC_r - nC_0 \]

### Step 4: Substituting the known values

From the binomial theorem, we have:

\[ \sum_{r=0}^{n} nC_r = 2^n \]

And since \(nC_0 = 1\), we can substitute these values into our equation:

\[ \sum_{r=1}^{n} nC_r = 2^n - 1 \]

### Conclusion

Thus, the value of the original summation is:

\[ \sum_{r=1}^{n} \frac{nP_r}{r!} = 2^n - 1 \]

### Final Answer

The final answer is:

\[ 2^n - 1 \]

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