Rewrite them in terms of combination maybe?

$\dfrac{6!\times 7!}{5!\times 8!} = \dfrac{6}{8} < 1$

$\dfrac{6!\times 7!}{4!\times 9!} = \dfrac{30}{72} < 1$

Hence $6!\times 7!$ is indeed the smallest.

Let $\color{#D61F06}a = 6!7!$ , $\color{#3D99F6}b = 5!8!$ and $\color{#20A900}c = 4!9!$ . Observe that $\begin{array}{rl} 6!9! &= 6! \cdot (9\cdot 8) 7! = 72 \cdot \left({\color{#D61F06}6!7!}\right) = 72{\color{#D61F06}a}\\ &= \left(6 \cdot 5!\right) \cdot \left(9 \cdot 8!\right) = 6 \cdot 9 \cdot \left({\color{#3D99F6}5!8!}\right) = 54{\color{#3D99F6}b}\\ &= \left(6 \cdot 5 \cdot 4! \right) \cdot 9! = 6 \cdot 5 \cdot \left({\color{#20A900}4!9!}\right) = 30{\color{#20A900}c} \end{array}$ Thus, $\begin{array}{rrcl} & 72{\color{#D61F06}a} &= 54{\color{#3D99F6}b} &= 30{\color{#20A900}c}\\ & {\color{#D61F06}a} &= \dfrac{3}{4}{\color{#3D99F6}b} &= \dfrac{5}{12}{\color{#20A900}c}\\ \Longrightarrow& {\color{#D61F06}a} &< {\color{#3D99F6}b} &< {\color{#20A900}c} \end{array}$ which shows that $\boxed{{\color{#D61F06}6!7!}\text{ is the smallest}}$ .

Let $a,b,c$ denote the values $4! \times9!, 5!\times8!, 6!\times7!$ , respectively.

We want to find $\min(a,b,c)$ . Equivalently, we can find $\max \left( \dfrac1a, \dfrac1b, \dfrac1c \right)$ .

Or similarly, $\max \left( \dfrac{13!}a, \dfrac{13!}b, \dfrac{13!}c \right)$ .

Since $\dfrac{13!}a = ^{13}C_4, \dfrac{13!}b = ^{13}C_5, \dfrac{13!}c =^{13}C_6$ , where $^a C_b = \dfrac{a!}{b!(a-b)!}$ denotes the binomial coefficient .

Then, $\max \left( \dfrac{13!}a, \dfrac{13!}b, \dfrac{13!}c \right) = \max \left(^{13}C_4, ^{13}C_5, ^{13}C_6\right)$ .

Using the properties of binomial coefficient, we know that $^{13}C_x$ is maximized when $x = \left \lfloor \dfrac{13}2 \right \rfloor , \left \lceil \dfrac{13}2 \right \rceil = 6, 7$ . So $\max \left(^{13}C_4, ^{13}C_5, ^{13}C_6\right) = ^{13}C_6 = \dfrac{13!}{c}$ . Hence, the smallest numbers among $a,b,c$ is $c = \boxed{6!\times7!}$ .

Recall that, for a fixed natural number $n$ , the binomial coefficient $\binom{n}{k}= \frac{n!}{k! (n-k)!}$ attains its maximum value when $k=\lfloor \frac{n}{2}\rfloor$ . This implies that, for a fixed $n$ , the product $k!(n-k)!$ attains its minimum value for $k=\lfloor \frac{n}{2}\rfloor$ . Apply the above result with $n=13$ .

4!9! = 4!4! * (5 * 6 * 7 * 8 * 9) = 4!4! * (5 * 6 * 7) * (8 * 9)
5!8! = 4!4! * (5 * 5 * 6 * 7 * 8) = 4!4! * (5 * 6 * 7) * (5 * 8)
6!7! = 4!4! * (5 * 6 * 5 * 6 * 7) = 4!4! * (5 * 6 * 7) * (5 * 6)

(5 * 6)<(5 * 8)<(8 * 9)

This is how do I mentally solved the problem.

When a×b is min when a approaches b