Question

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

• A. 70
• B. 60
• C. 50
• D. None of these

Answer 1

Answer: (B)

We have
$(xy+yz+zx{)}^6={\sum }_{r+s+t=6}\frac{6!}{r!s!t!}(xy{)}^r(yz{)}^s(zx{)}^t$
$={\sum }_{r+s+t=6}\frac{6!}{r!s!t!}{x}^{r+t}{y}^{r+s}{z}^{s+t}$
If the general term in the above expansion.
contains $x^3{y}^4{z}^5$, then
$r+t=3,r+s=4$ and $s+t=5$
Also, $r+s+t=6$
Solving these equations, we get
$r=1,s=3,t=2$
$\therefore$ Coefficient of
$x^3{y}^4{z}^5=\frac{6!}{1!3!2!}=\frac{6!}{2!3!}=60$