The coefficient of x^3 y^4 z^5 in the expansi | vedclass

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(B) We have the multinomial expansion formula: $(x y+y z+z x)^6 = \sum_{r+s+t=6} \frac{6!}{r! s! t!} (x y)^r (y z)^s (z x)^t$ $= \sum_{r+s+t=6} \frac{

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(B) We have the multinomial expansion formula: $(x y+y z+z x)^6 = \sum_{r+s+t=6} \frac{6!}{r! s! t!} (x y)^r (y z)^s (z x)^t$ $= \sum_{r+s+t=6} \frac{6!}{r! s! t!} x^{r+t} y^{r+s} z^{s+t}$ For the term $x^3 y^4 z^5$,we equate the exponents: $r+t=3$ $r+s=4$ $s+t=5$ Adding these three equations: $2(r+s+t) = 12 \implies r+s+t = 6$. Subtracting the equations from the sum: $s = (r+s+t) - (r+t) = 6 - 3 = 3$ $t = (r+s+t) - (r+s) = 6 - 4 = 2$ $r = (r+s+t) - (s+t) = 6 - 5 = 1$ Thus,the coefficient is $\frac{6!}{1! 3! 2!} = \frac{720}{1 	imes 6 	imes 2} = 60$.

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

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