Question

Find the coefficient of $a^3{b}^4{c}^5$ in ${(ab+bc+ca)}^6$.

• A. $60$
• B. $45$
• C. $40$
• D. $90$

Answer 1

The correct option is A

$60$

Explanation for the correct option:

Step:1 Simplify the series and expand it to get the coefficient

Let, $p,q,$ and $r$ are three numbers.

The coefficient of $a^3{b}^4{c}^5$ in

$\begin{array}{rcl}{(ab+bc+ca)}^6& =& \sum _{p+q+r=6}\frac{6!}{p!q!r!}{\left(ab\right)}^p{\left(bc\right)}^q{\left(ca\right)}^r\\ & =& \sum _{p+q+r=6}\frac{6!}{p!q!r!}{\left(a\right)}^{p+r}{\left(b\right)}^{p+q}{\left(a\right)}^{q+r}\end{array}$.

We can write,

$$\begin{gathered} p+r=3 \\ p+q=4 \\ q+r=5 \end{gathered}$$

Therefore, $\therefore p=1,q=3,r=2$.

Srep2. Finding coefficient.

The coefficient of $a^3{b}^4{c}^5$ in ${(ab+bc+ca)}^6$.is

$$\begin{gathered} =\frac{6!}{1!3!2!} \\ =\frac{6\times 5\times 4\times 3!}{2\times 3!} \\ =60 \end{gathered}$$

Hence, Option (A) is the correct answer.