We aim to evaluate the given summation: \[ \sum_{r=1}^n \frac{1}{2^n} \cdot \frac{nPr}{r!}. \] Simplify \(nPr\), The formula for \(nPr\) is: \[ nPr = \frac{n!}{(n-r)!}. \] Substituting this into the summation gives: \[ \sum_{r=1}^n \frac{1}{2^n} \cdot \frac{\frac{n!}{(n-r)!}}{r!}. \] Simplify the fraction: \[ \sum_{r=1}^n \frac{n!}{2^n \cdot r! \cdot (n-r)!}. \] Recognize that this is the binomial coefficient: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!}. \] Thus, the summation becomes: \[ \sum_{r=1}^n \frac{\binom{n}{r}}{2^n}. \] The binomial theorem states: \[ \sum_{r=0}^n \binom{n}{r} \cdot x^r \cdot y^{n-r} = (x + y)^n. \] Here, set \(x = \frac{1}{2}\) and \(y = \frac{1}{2}\), so that: \[ \sum_{r=0}^n \binom{n}{r} \cdot \left(\frac{1}{2}\right)^r \cdot \left(\frac{1}{2}\right)^{n-r} = \left(\frac{1}{2} + \frac{1}{2}\right)^n = 1^n = 1. \] Thus: \[ \sum_{r=0}^n \frac{\binom{n}{r}}{2^n} = 1. \] The summation starts from \(r = 1\), so we exclude the term for \(r = 0\): \[ \sum_{r=1}^n \frac{\binom{n}{r}}{2^n} = 1 - \frac{\binom{n}{0}}{2^n}. \] Since \(\binom{n}{0} = 1\), this becomes: \[ \sum_{r=1}^n \frac{\binom{n}{r}}{2^n} = 1 - \frac{1}{2^n}. \] \[{1 - 2^{-n}} \]