Q49 / 160 · TS EAMCET 2005 · Mathematics

The coefficient of $x^3 y^4 z^5$ in the expansion of $(x y+y z+x z)^6$ is

  • A
    70
  • Correct answer
    B
    60
  • C
    50
  • D
    None of these

Solution

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

Hence, the correct answer is option: 'B'.

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