The coefficient of \[{{a}^{8}}{{b}^{4}}{{c}^{9}}{{d}^{9}}\] in \[{{\left( abc+abd+acd+bcd \right)}^{10}}\] is
(A) \[10!\]
(B) \[\dfrac{10!}{8!4!9!9!}\]
(C) 2520
(D) none of these.

Hint: The general term of the expression \[{{\left( a+b+c+d \right)}^{10}}\] is given by the formula \[\dfrac{10!}{p!q!r!s!}{{a}^{p}}{{b}^{q}}{{c}^{r}}{{d}^{s}}\] . Now, replace \[a\] by \[abc\] , \[b\] by \[abd\] , \[c\] by \[acd\] , and \[d\] by \[bcd\] in the formula. Now, get the exponents of a, b, c, and d from the expression \[\dfrac{10!}{p!q!r!s!}{{\left( abc \right)}^{p}}{{\left( abd \right)}^{q}}{{\left( acd \right)}^{r}}{{\left( bcd \right)}^{s}}\] and then compare with the exponents of a, b, c, and d from the expression \[{{a}^{8}}{{b}^{4}}{{c}^{9}}{{d}^{9}}\] . Now, solve it further and get the value of p, q, r, and s. Then, put the value of p, q, r, and s in the expression \[\dfrac{10!}{p!q!r!s!}{{a}^{p}}{{b}^{q}}{{c}^{r}}{{d}^{s}}\] and get the coefficient of \[{{a}^{8}}{{b}^{4}}{{c}^{9}}{{d}^{9}}\].

Complete step by step solution:
According to the question, our given expression is \[{{\left( abc+abd+acd+bcd \right)}^{10}}\].
We know the formula that the general term of the expression \[{{\left( a+b+c+d \right)}^{10}}\] is given by
\[\dfrac{10!}{p!q!r!s!}{{a}^{p}}{{b}^{q}}{{c}^{r}}{{d}^{s}}\] ……………….(1)
Now, replacing \[a\] by \[abc\] , \[b\] by \[abd\] , \[c\] by \[acd\] , and \[d\] by \[bcd\] in equation (1), we get,
The general term of the expression \[{{\left( abc+abd+acd+bcd \right)}^{10}}\] is given by \[\dfrac{10!}{p!q!r!s!}{{\left( abc \right)}^{p}}{{\left( abd \right)}^{q}}{{\left( acd \right)}^{r}}{{\left( bcd \right)}^{s}}\] ………………(2)
Simplifying equation (2), we get
\[\begin{align}
  & \dfrac{10!}{p!q!r!s!}{{\left( abc \right)}^{p}}{{\left( abd \right)}^{q}}{{\left( acd \right)}^{r}}{{\left( bcd \right)}^{s}} \\
 & =\dfrac{10!}{p!q!r!s!}{{\left( a \right)}^{p}}{{\left( a \right)}^{q}}{{\left( a \right)}^{r}}{{\left( b \right)}^{p}}{{\left( b \right)}^{q}}{{\left( b \right)}^{s}}{{\left( c \right)}^{p}}{{\left( c \right)}^{r}}{{\left( c \right)}^{s}}{{\left( d \right)}^{q}}{{\left( d \right)}^{r}}{{\left( d \right)}^{s}} \\
 & =\dfrac{10!}{p!q!r!s!}{{\left( a \right)}^{\left( p+q+r \right)}}{{\left( b \right)}^{\left( p+q+s \right)}}{{\left( c \right)}^{\left( p+r+s \right)}}{{\left( d \right)}^{\left( q+r+s \right)}} \\
\end{align}\]
So, the general term of \[{{\left( abc+abd+acd+bcd \right)}^{10}}\] is \[\dfrac{10!}{p!q!r!s!}{{\left( a \right)}^{\left( p+q+r \right)}}{{\left( b \right)}^{\left( p+q+s \right)}}{{\left( c \right)}^{\left( p+r+s \right)}}{{\left( d \right)}^{\left( q+r+s \right)}}\] …………………..(3)
It is given that we have to find the coefficient of \[{{a}^{8}}{{b}^{4}}{{c}^{9}}{{d}^{9}}\] ………………(4).
So, we have to compare the exponents of a, b, c, and d of equation (3) and equation (4).
On comparing the exponents of a, b, c, and d of equation (3) and equation (4), we get
\[p+q+r=8\] ………………(5)
\[p+q+s=4\] ……………….(6)
\[p+r+s=9\] ………………..(7)
\[q+r+s=9\] ………………..(8)
Subtracting equation (7) from equation (8), we get
\[\begin{align}
  & \left( q+r+s \right)-\left( p+r+s \right)=9-9 \\
 & \Rightarrow q-p=0 \\
\end{align}\]
\[\Rightarrow q=p\] ………………………(9)
Now, subtracting equation (6) from equation (5)
\[\begin{align}
  & \left( p+q+r \right)-\left( p+q+s \right)=8-4 \\
 & \Rightarrow r-s=4 \\
\end{align}\]
\[\Rightarrow r=s+4\] …………………….(10)
Now, subtracting equation (5) from equation (8), we get
\[\begin{align}
  & \left( q+r+s \right)-\left( p+q+r \right)=9-8 \\
 & \Rightarrow s-p=1 \\
\end{align}\]
\[\Rightarrow s=p+1\] ……………….(11)
Putting the value of s from equation (11), in equation (10), we get
\[\begin{align}
  & \Rightarrow r=s+4 \\
 & \Rightarrow r=p+1+4 \\
\end{align}\]
\[\Rightarrow r=p+5\] ………………………(12)
Putting the value of q and r from equation (9) and equation (12), in equation (5), we get
From equation (5), equation (9), and equation (12), we have,
\[\begin{align}
  & p+q+r=8 \\
 & \Rightarrow p+p+p+5=8 \\
 & \Rightarrow 3p=3 \\
 & \Rightarrow p=1 \\
\end{align}\]
Now, put the value of p in equation (9), we get
\[\begin{align}
  & q=p \\
 & \Rightarrow q=1 \\
\end{align}\]
Now, put the value of p in equation (11), we get
\[\begin{align}
  & s=p+1 \\
 & \Rightarrow s=1+1 \\
 & \Rightarrow s=2 \\
\end{align}\]
Now, put the value of p in equation (12), we get
\[\begin{align}
  & r=p+5 \\
 & \Rightarrow r=1+5 \\
 & \Rightarrow r=6 \\
\end{align}\]
So, the value of p, q, r, and s is 1, 1, 6, and 2.
Now, putting the value of p, q, r, and s in equation (3), we get
\[\begin{align}
  & \dfrac{10!}{p!q!r!s!}{{\left( a \right)}^{\left( p+q+r \right)}}{{\left( b \right)}^{\left( p+q+s \right)}}{{\left( c \right)}^{\left( p+r+s \right)}}{{\left( d \right)}^{\left( q+r+s \right)}} \\
 & =\dfrac{10!}{1!1!6!2!}{{\left( a \right)}^{\left( 1+2+5 \right)}}{{\left( b \right)}^{\left( 1+1+2 \right)}}{{\left( c \right)}^{\left( 1+6+2 \right)}}{{\left( d \right)}^{\left( 1+6+2 \right)}} \\
 & =\dfrac{10!}{1!1!6!2!}{{\left( a \right)}^{8}}{{\left( b \right)}^{4}}{{\left( c \right)}^{9}}{{\left( d \right)}^{9}} \\
 & =\dfrac{\left( 6! \right)\times 7\times 8\times 9\times 10}{6!2!}{{\left( a \right)}^{8}}{{\left( b \right)}^{4}}{{\left( c \right)}^{9}}{{\left( d \right)}^{9}} \\
 & =7\times 4\times 9\times 10{{\left( a \right)}^{8}}{{\left( b \right)}^{4}}{{\left( c \right)}^{9}}{{\left( d \right)}^{9}} \\
 & =2520{{\left( a \right)}^{8}}{{\left( b \right)}^{4}}{{\left( c \right)}^{9}}{{\left( d \right)}^{9}} \\
\end{align}\]
Therefore, the coefficient of \[{{\left( a \right)}^{8}}{{\left( b \right)}^{4}}{{\left( c \right)}^{9}}{{\left( d \right)}^{9}}\] is 2520.
Hence, the correct option is 2520.

Note: To solve this question, one may think to take a as common on whole make the expression\[{{\left( abc+abd+acd+bcd \right)}^{10}}\] as \[{{a}^{10}}{{b}^{10}}{{c}^{10}}{{\left( 1+\dfrac{abd+acd+bcd}{abc} \right)}^{10}}\] and then think to apply the binomial expansion of \[{{\left( 1+x \right)}^{n}}\] . If we do so then our expression will become complex. So, here formula should be used in order to reduce the complexity.