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If \( P(n): 2^{n} < n ! \) then the smallest positive integer for which \( P(n) \) is true, is
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The correct answer is:
\( 04 \)
$\$ P(\mathrm{n}): 2^{\wedge}\{\mathrm{n}\}$ Since
$P(1): 2 < 1$ is false.
$P(2): 2^{2} < 2 !$ is false
$P(3): 2^{3} < 3 !$ is false
$P(4): 2^{4} < 4 !$ is true.
$P(5): 2^{5} < 5 !$ is true.
So, the smallest positive integer is $4 .$
$P(1): 2 < 1$ is false.
$P(2): 2^{2} < 2 !$ is false
$P(3): 2^{3} < 3 !$ is false
$P(4): 2^{4} < 4 !$ is true.
$P(5): 2^{5} < 5 !$ is true.
So, the smallest positive integer is $4 .$
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