# Probability Probelms & Counting Techniques

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• Jun 10th 2008, 03:43 PM
16sammy
Probability Probelms & Counting Techniques
Expand using the binomial theorem. Simplify the result fully.

(x^2 - 3y)^4
• Jun 10th 2008, 04:27 PM
Nyoxis
$\displaystyle (x^2)^4 + 4\{(x^2)^3 \cdot (-3y)^1\} + 6\{(x^2)^2 \cdot (-3y)^2\} + 4\{(x^2)^1 \cdot (-3y)^3\} + (-3y)^4 \\$

$\displaystyle \Rightarrow x^8 + 4(x^6 \cdot -3y) +6(x^4 \cdot 9y^2) + 4(x^2 \cdot -27y^3) + 81y^4$

...

Ouch, I'm pretty tired- thanks Soroban.
• Jun 10th 2008, 04:58 PM
Soroban
Hello, 16sammy

Quote:

Expand using the binomial theorem: .$\displaystyle (x^2 - 3y)^4$
We have: .$\displaystyle (x^2-3y)^4$

. . $\displaystyle = \;\;(x^2)^4 \;+ \;4(x^2)^3(\text{-}3y)^1 \;+\; 6(x^2)^2(\text{-}3y)^2 \;+\; 4(x^2)^1(\text{-}3y)^3 \;+\; (\text{-}3y)^4$

. . $\displaystyle = \;\;x^8 \;+\; 4(x^6)(\text{-}3y) \;+ \;6(x^4)(9y^2) \;+ \;4(x^2)(\text{-}27y^3) \;+ \;81y^4$

. . $\displaystyle = \;\;x^8 \;- \;12x^6y \;+ \;54x^4y^2 \;- \;108x^2y^3 \;+ \;81y^4$