Question

What is the coefficient of $a^3{b}^4{c}^5$ in the expansion of ${(ab+bc+ca)}^6$?

Answer 1

Let us consider coefficient of $x^p{y}^q{r}^{n-p-q}$ in the expansion of ${(x+y+z)}^n$

We have $p\ge 0,n\ge p+q.$

Now we can get $p$ in ${}^n{C}_p$ ways and from remaining $(n-p)$ ways we can get $q$ in ${}^{n-p}{C}_q$

Hence $x^p{y}^q{z}^r$ can occur in $\frac{n!}{p!q!(n-p-q)!}$ ways.

Consider ${(ab+bc+ca)}^6$

$={\left(abc\right)}^6{\left(\frac{ab+bc+ca}{abc}\right)}^6$

$={\left(abc\right)}^6{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}^6$

And we are seeking coefficient of $a^3{b}^4{c}^5$

$={\left(abc\right)}^6{\left(\frac{1}{a}\right)}^3{\left(\frac{1}{b}\right)}^2\left(\frac{1}{c}\right)$

If $x=\frac{1}{a},y=\frac{1}{b}$ and $z=\frac{1}{c}$

Coefficent of $x^3{y}^{2}z$ in ${(x+y+z)}^6$

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