Solution using $10$:
$\big((10!!)_!\big)_!-\frac{10}{\Big(\big((10!!)_!\big)_!\Big)_!} = (3840_!)_!-\frac{10}{\big((3840_!)_!\big)_!} = 38_!-\frac{10}{(38_!)_!} = 11-\frac{10}{11_!} = 11-\frac{10}{2} = 11-5=6$ where $(x)_!$ is the factorial number system.
Another solution using $10$ which I don't regard as an answer though (a silly one) is:
$10+10+10 = 6$ (binary $2+2+2=6$)
Solution using two $9$s:
$$\Bigg(\frac{(9!!)_!}{9}\Bigg)! = \Bigg(\frac{945_!}{9}\Bigg)!=\Bigg(\frac{27}{9}\Bigg)!=6$$
If we are allowed to use another mathematical operator $\%$ (modulus, a close relative of division) then the solution using three $9$s is:
$$\Bigg(\frac{(9!!)_!}{9}\Bigg)!\%9 = \Bigg(\frac{945_!}{9}\Bigg)!\%9=\Bigg(\frac{27}{9}\Bigg)!\%9=6\%9=6$$
Solution using just one mathematical operator $\bar{}$ ($1$'s complement) if allowed using $9$
$\bar{9} = \overline{1001} = 0110 = 6$
Leading to three $9$s solution using two operations:
$\bar{9}+9-9 = 6$
Also, an interesting thing to note is that
$6 = 9$ (XS3 code) or using three $9$s as exprected $6=9+9-9 $
Another solution using $10$
$\Big(-\big(-(10!!)_{-10}\big)_{-10}\Big)_{-!} = \Big(-\big(-3840_{-10}\big)_{-10}\Big)_{-!}= \Big(-2240_{-10}\Big)_{-!} = 1840_{-!} = 1\times (-1)^3\times 3! + 8\times (-1)^2 \times 2! + 4\times (-1)^1\times 1! = 6$ where $(x)_{-10}$ is the negadecimal system and $(x)_{-!}$ is the negative factorial system.