  Question

Let $$P(n)$$ be the statement $$2^{n}<n!$$ where $$n$$ is a natural number, then $$P(n)$$ is true for:

A
all n  B
all n>2  C
all n>3  D
all n<3  Solution

The correct option is C all $$n>3$$We have, $$P\left( n \right)$$ be the statement $${{2}^{n}}<n!$$ Where $$n$$ is a natural number Then, Put $$n=1,2,3....$$ So, $$P\left( 1 \right)\,$$ be the statement of $${{2}^{1}}<1!=2<1\,\,\,\left( \text{It}\,\text{is}\,\text{wrong} \right)$$ $$P\left( 2 \right)$$  be the statement of $${{2}^{2}}<2!=4<2\,\,\left( \text{It}\,\text{is}\,\text{wrong} \right)$$$$P\left( 3 \right)$$  be the statement of $${{2}^{3}}<3!=8<6\,\,\left( \text{It}\,\text{is}\,\text{wrong} \right)$$But,$$P\left( 4 \right)$$  be the statement of $${{2}^{4}}<4!=16<24\,\,\left( \text{It}\,\text{is}\,\text{right} \right)$$ Similarly, $$P\left( 5 \right),\,P\left( 6 \right)\,.......$$ So, all number $$n$$ is a natural number.Therefore, it is true for all $$n>3$$.Hence, this is the answer.Maths

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