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

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

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$60$

Vedclass · Answer

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

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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