Let \( P_{m} \) stand for \( { }^{m} P_{m} \), then, \( 1+P_{1}+2 P_{2}+3 P_{3}+\ldots+n \cdot P...

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Let \( P_{m} \) stand for \( { }^{m} P_{m} \), then, \( 1+P_{1}+2 P_{2}+3 P_{3}+\ldots+n \cdot P_{n} \) is equal to
(1) \( (n-1) \) !
(2) \( n \) !
(3) \( (n+1) !-1 \)
(4) \( (n+1) \) !
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