+537 Find the coefficient of x^3 y^3 z^2 in the expansion of (x+y+z)^8.
+26393 Find the coefficient of x^3 y^3 z^2 in the expansion of (x+y+z)^8.
Trinomial Coefficient:
\(\begin{array}{|rcll|} \hline \dbinom{8}{3,3,2} &=& \dfrac{8!}{3!3!2!} \\\\ &=& \dfrac{3!4\cdot 5 \cdot 6 \cdot 7 \cdot 8}{3!3!2!} \\\\ &=& \dfrac{4\cdot 5 \cdot 6 \cdot 7 \cdot 8}{3!2!} \quad & | \quad 3! = 6 \quad 2! = 2 \\\\ &=& \dfrac{4\cdot 5 \cdot 6 \cdot 7 \cdot 8}{6\cdot 2} \\\\ &=& \dfrac{4\cdot 5 \cdot 7 \cdot 8}{2} \\\\ &=& 2\cdot 5 \cdot 7 \cdot 8 \\\\ &=& 10 \cdot 56 \\\\ &\mathbf{=}& \mathbf{ 560 } \\ \hline \end{array}\)
The trinomial coefficient of \(x^3 y^3 z^2\) in the expansion of \((x+y+z)^8\) is 560
+118677 Thanks Heureka, I can't remember ever seeing this type of solution before.
So I am very pleased that you have shown me.
+118677 Hi Heureka,
I am wondering if it can still be done easily if it is made more complicated ?
eg
Find the coefficient of x^3 y^3 z^2 in the expansion of (3x+5y+7z)^8.
+26393 Hi Melody
"Find the coefficient of x^3 y^3 z^2 in the expansion of (3x+5y+7z)^8".
Set \(a = 3x\)
Set \(b = 5y\)
Set \(c = 7z\)
Set \(n = 8\)
Set \(i = 3\)
Set \(j = 3\)
Set \(k = 2\)
Trinomial Coefficient of \(x^3 y^3 z^2\):
\(\begin{array}{|rcll|} \hline \dbinom{8}{3,3,2}(3x)^3\cdot(5y)^3\cdot (7z)^2 &=& \dfrac{8!}{3!3!2!} (3x)^3\cdot(5y)^3\cdot (7z)^2 \\\\ &=& \dfrac{8!}{3!3!2!} \cdot 3^35^37^2 x^3y^3z^2 \quad & | \quad \dfrac{8!}{3!3!2!} = 560 \\\\ &=& 560\cdot 3^35^37^2 x^3y^3z^2 \\\\ &=& 560\cdot 27 \cdot 125 \cdot 49 x^3y^3z^2 \\\\ &\mathbf{=}& \mathbf{ 92610000 x^3y^3z^2 } \\ \hline \end{array}\)
\(\text{The coefficient of $x^3 y^3 z^2$ in the expansion of $(3x+5y+7z)^8$ is $\mathbf{92~ 610~ 000}$ } \)
Source: https://en.wikipedia.org/wiki/Trinomial_expansion
