Question

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.

Answer 1

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

& \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{aligned}$$ So, the general term of $${{\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{aligned} & \left( q+r+s \right)-\left( p+r+s \right)=9-9 \\\ & \Rightarrow q-p=0 \\\ \end{aligned}$$ $$\Rightarrow q=p$$ ………………………(9) Now, subtracting equation (6) from equation (5) $$\begin{aligned} & \left( p+q+r \right)-\left( p+q+s \right)=8-4 \\\ & \Rightarrow r-s=4 \\\ \end{aligned}$$ $$\Rightarrow r=s+4$$ …………………….(10) Now, subtracting equation (5) from equation (8), we get $$\begin{aligned} & \left( q+r+s \right)-\left( p+q+r \right)=9-8 \\\ & \Rightarrow s-p=1 \\\ \end{aligned}$$ $$\Rightarrow s=p+1$$ ……………….(11) Putting the value of s from equation (11), in equation (10), we get $$\begin{aligned} & \Rightarrow r=s+4 \\\ & \Rightarrow r=p+1+4 \\\ \end{aligned}$$ $$\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{aligned} & p+q+r=8 \\\ & \Rightarrow p+p+p+5=8 \\\ & \Rightarrow 3p=3 \\\ & \Rightarrow p=1 \\\ \end{aligned}$$ Now, put the value of p in equation (9), we get $$\begin{aligned} & q=p \\\ & \Rightarrow q=1 \\\ \end{aligned}$$ Now, put the value of p in equation (11), we get $$\begin{aligned} & s=p+1 \\\ & \Rightarrow s=1+1 \\\ & \Rightarrow s=2 \\\ \end{aligned}$$ Now, put the value of p in equation (12), we get $$\begin{aligned} & r=p+5 \\\ & \Rightarrow r=1+5 \\\ & \Rightarrow r=6 \\\ \end{aligned}$$ So, the value of p, q, r, and s is 1, 1, 6, and 2. Now, putting the value of p, q, r, and s in equation (3), we get $$\begin{aligned} & \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{aligned}$$ Therefore, the coefficient of $${{\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.