The value of $\sum_{r=1}^n \frac{{ }^n P_r}{r !}$ is
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The value of $\sum_{r=1}^n \frac{{ }^n P_r}{r !}$ is

(A) $2^n$

(B) $2^n-1$

(C) $2^{n-1}$

(D) $2^n+1$

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SOLUTION :

$\therefore \sum_{r=1}^n \frac{{ }^n P_r}{r !}=\sum_{r=1}^n{ }^n C_r$

$={ }^n C_1+{ }^n C_2+\ldots+{ }^n C_n=2^n-1$

So, The correct option is (B).

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