The value of $\sum_{r=1}^{n} (1/2^n) ({}^nP_r)/r!$ is
- A. 2^n
- B. 1 - 2^{-n}
- C. 2^n - 1
- D. 2^{2n} - 1
✅ Correct Answer: 1 - 2^{-n}
Explanation
$({}^nP_r)/r! = {}^nC_r$ So the sum becomes $(1/2^n) \sum_{r=1}^{n} {}^nC_r$ $= (1/2^n)(2^n - 1)$ $= 1 - 2^{-n}$
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