Perfect Square

All of the products are of the form $n! \cdot (n + 1)!$ or $n! \cdot (n + 2)!$ . Observe that

$\displaystyle n! \cdot (n + 1)! = (n!)^2 \cdot (n + 1)$

$\displaystyle n! \cdot (n + 2)! = (n!)^2 \cdot (n + 1)(n + 2)$

Thus, each product $p$ is a perfect square if and only if $\displaystyle \frac{p}{(n!)^2}$ is also a perfect square.

Only $\boxed{B}$ fits that criteria:

$B = 99! \cdot 100! = (99!)^2 \cdot 100$

Note that $a^2 \cdot b^2 = (ab)^2 = m^2$ for some integer $m$ .

A. $98!\times99! = 98!\times 99\times 98! = (98!)^2\times 99 \implies$ not a perfect square.

B. $99!\times100! = 99!\times 100\times 99! = (99!)^2\times 100 = (99!\times 10)^2 \implies$ a perfect square .

C. $98!\times100! = 98!\times 100\times 99\times 98! = (98!)^2\times 100 \times 99 \implies$ not a perfect square.

D. $100!\times101! = 100!\times 101\times 100! = (100!)^2\times 101 \implies$ not a perfect square.

E. $99!\times101! = 99!\times 101\times 100\times 99! = (99!)^2\times 101\times 100 \implies$ not a perfect square.

$= n! \cdot (n + 1)!$

$= n! \cdot n!(n+1)$

$= (n!)^2 \cdot (n + 1)$

$\sqrt { { (n!) }^{ 2 }(n+1) } =\sqrt { { (n!) }^{ 2 } } \cdot \sqrt { n+1 }$ is integer.

So, the answer is $99! \times 100!$

Only B, $99! \times 100! = (99!)^2 \times 100 = 99!^2 \times 10^2$