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# Binomial Theroem

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Find the coefficient of x^3 y^3 z^2 in the expansion of (x+y+z)^8.

Aug 9, 2018

#1
+21295
+4

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

Aug 9, 2018
edited by heureka  Aug 9, 2018
#2
+97561
+1

Thanks Heureka, I can't remember ever seeing this type of solution before.

So I am very pleased that you have shown me.

Aug 9, 2018
#3
+21295
+2

Thank you

heureka  Aug 9, 2018
#4
+97561
0

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.

Aug 9, 2018
#5
+21295
+3

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  $$x^3 y^3 z^2$$:
$$\begin{array}{|rcll|} \hline \dbinom{8}{3,3,2}(3x)^3\cdot(5y)^3\cdot (7z)^2 &=& \dfrac{8!}{3!3!2!} (3x)^3\cdot(5y)^3\cdot (7z)^2 \\\\ &=& \dfrac{8!}{3!3!2!} \cdot 3^35^37^2 x^3y^3z^2 \quad & | \quad \dfrac{8!}{3!3!2!} = 560 \\\\ &=& 560\cdot 3^35^37^2 x^3y^3z^2 \\\\ &=& 560\cdot 27 \cdot 125 \cdot 49 x^3y^3z^2 \\\\ &\mathbf{=}& \mathbf{ 92610000 x^3y^3z^2 } \\ \hline \end{array}$$

$$\text{The coefficient of x^3 y^3 z^2 in the expansion of (3x+5y+7z)^8 is \mathbf{92~ 610~ 000} }$$

heureka  Aug 10, 2018
edited by heureka  Aug 10, 2018
#6
+97561
+2

Thanks Heureka. That is great !!

Melody  Aug 10, 2018