Question

The coefficient of $x^3{y}^4{z}^5$ in the expansion $(xy+yz+xz{)}^6$ is

• A. $60$
• B. $50$
• C. $40$
• D. $70$

Answer 1

The correct option is A $60$

The expansion can be written as
$(xy+yz+xz{)}^6=\sum _{r+s+t=6}\frac{6!}{(r!)(s!)(t!)}\left(xy{)}^r\right(yz{)}^s(xz{)}^t=\sum _{r+s+t=6}\frac{6!}{(r!)(s!)(t!)}\left(x{)}^{r+t}\right(y{)}^{r+s}(z{)}^{s+t}$

Where, $0\le r,s,t\le 6$
Comparing the coefficients of $\left(x{)}^{r+t}\right(y{)}^{r+s}(z{)}^{s+t}$ with $x^3{y}^4{z}^5$, we get
$r+s+t=6r+t=3\Rightarrow s=3r+s=4\Rightarrow t=2\therefore r=1,s=3,t=2$

So, the coefficient of $x^3{y}^4{z}^5$
$=\frac{6!}{(1!)(3!)(2!)}=\frac{6\cdot 5\cdot 4}{2}=60$