In the expansion of (x-2y+3z)^5,if the total | vedclass

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(B) The number of terms in the expansion of $(x+y+z)^n$ is given by the formula $\frac{(n+1)(n+2)}{2}$ or $^{n+k-1}C_{k-1}$,where $n=5$ and $k=3$.Tota

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$-\frac{180}{7}$

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(B) The number of terms in the expansion of $(x+y+z)^n$ is given by the formula $\frac{(n+1)(n+2)}{2}$ or $^{n+k-1}C_{k-1}$,where $n=5$ and $k=3$.Total number of terms $p = {}^{5+3-1}C_{3-1} = {}^{7}C_{2} = \frac{7 	imes 6}{2} = 21$.Thus,$p = 21$.Using the multinomial theorem,the coefficient of $x^a y^b z^c$ in $(x+y+z)^n$ is $\frac{n!}{a!b!c!} (coeff_x)^a (coeff_y)^b (coeff_z)^c$.For $x^2 y^1 z^2$ in $(x-2y+3z)^5$,the coefficient $q$ is $\frac{5!}{2!1!2!} (1)^2 (-2)^1 (3)^2$.$q = \frac{120}{4} 	imes (-2) 	imes 9 = 30 	imes (-18) = -540$.Therefore,$\frac{q}{p} = \frac{-540}{21} = -\frac{180}{7}$.

In the expansion of $(x-2y+3z)^5$,if the total number of terms is $p$ and the coefficient of $x^2yz^2$ is $q$,then $\frac{q}{p}=$

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