-Factorial-

1!<1.5!<2!,1<1.5!<2,so the answer is 1

$\displaystyle \lfloor { \frac { 3 } { 2 } ! } \rfloor = 1$ .

Obviously this question goes beyond the definition of the usual factorial function.

$n!=n(n-1)!$

$\frac { 3 }{ 2 } !=\frac { 3 }{ 2 } (\frac { 3 }{ 2 } -1)!$

$\frac { 3 }{ 2 } !=\frac { 3 }{ 2 } (\frac { 1 }{ 2 } )!$

We know that $n!=\Pi (n)=\int _{ 0 }^{ \infty }{ { t }^{ n } } { e }^{ -t }dt$

So $\frac { 1 }{ 2 } !=\Pi (\frac { 1 }{ 2 } )=\int _{ 0 }^{ \infty }{ \sqrt { t } } { e }^{ -t }dt$

Solving for $\int _{ 0 }^{ \infty }{ \sqrt { t } } { e }^{ -t }dt$

we can let $u=\sqrt { t } ,\quad { u }^{ 2 }=t,\quad 2udu=dt$

We then get the integral $2\int _{ 0 }^{ \infty }{ { u }^{ 2 } } { e }^{ -{ u }^{ 2 } }du$

Integrating by parts we reach the result $\int _{ 0 }^{ \infty }{ { e }^{ -{ u }^{ 2 } } } du$

This is almost the famous Gaussian Integral but instead this goes from $u=0,\infty$

This leaves us with the result $(\frac { 1 }{ 2 } )!=\frac { \sqrt { \pi } }{ 2 }$

Multiplying this by $\frac { 3 }{ 2 }$ we get $\frac { 3\sqrt { \pi } }{ 4 }$

Therefore $\frac { 3 }{ 2 } !=\frac { 3\sqrt { \pi } }{ 4 }$

$\left\lfloor \frac { 3\sqrt { \pi } }{ 4 } \right\rfloor =\boxed{1}$