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Q.
The coefficient of $x^{3}\, y^{4}\, z^{5}$ in the expansion of $(x y+y z+x z)^{6}$ is
EAMCETEAMCET 2005
Solution:
We have
$(x y+y z+z x)^{6}=\sum_{r+s+t=6} \frac{6 !}{r ! s ! t !}(x y)^{r}(y z)^{s}(z x)^{t}$
$=\sum_{r+s+t=6} \frac{6 !}{r ! s ! t !} x^{r+t} y^{r+s} z^{s+t}$
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
$x^{3} \,y^{4}\, z^{5}=\frac{6 !}{1 ! 3 ! 2 !}=\frac{6 !}{2 ! 3 !}=60$