Guess I should call it a 'tetronomial'

$\\ { (abc+bcd+cda+dab) }^{ 10 }\\ =\quad { a }^{ 10 }{ b }^{ 10 }{ c }^{ 10 }{ d }^{ 10 }{ (\frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } +\frac { 1 }{ d } ) }^{ 10 }\\$

So the coefficient of ${ a }^{ 8 }{ b }^{ 4 }{ c }^{ 9 }{ d }^{ 9 }$ is simply the coefficient of ${ (\frac { 1 }{ a } })^{ 2 }{ (\frac { 1 }{ b } })^{ 6 }{ (\frac { 1 }{ c } })^{ 1 }{ (\frac { 1 }{ d } })^{ 1 }$ in the expansion ${ (\frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } +\frac { 1 }{ d } ) }^{ 10 }$

Hence by Multinomial theorem,

coefficient of ${ (\frac { 1 }{ a } })^{ 2 }{ (\frac { 1 }{ b } })^{ 6 }{ (\frac { 1 }{ c } })^{ 1 }{ (\frac { 1 }{ d } })^{ 1 }$ $=\quad \frac { 10! }{ 2!6!1!1! }$ $=\boxed { 2520 }$

$( abc + bcd + cda + dab )^{10}$

For ${a}^{m}{b}^{n}{c}^{o}{d}^{p}$ :

$= \frac {10!}{w!x!y!z!} \cdot{abc}^{w}\cdot{bcd}^{x}\cdot{cda}^{y}\cdot{dab}^{z}$

$= \frac {10!}{w!x!y!z!} \cdot{a}^{w+y+z}\cdot{b}^{w+x+z}\cdot{c}^{w+x+y}\cdot{d}^{x+y+z}$

where

$w+x+y+z = 10$

$m = w+y+z$

$n = w+x+z$

$o = w+x+y$

$p = x+y+z$

Solve for ${a}^{8}{b}^{4}{c}^{9}{d}^{9}$ , we have

$8 = w+y+z \rightarrow x = 2$

$4 = w+x+z \rightarrow y = 6$

$9 = w+x+y \rightarrow z = 1$

$9 = x+y+z \rightarrow w = 1$

Hence the coefficient of ${a}^{8}{b}^{4}{c}^{9}{d}^{9}$ :

$= \frac {10!}{1!2!6!1!}$

$= \boxed{2520}$