# How can I rewrite (w^3x^4y^6)^2 within parentheses without exponents?

Apr 10, 2017

See the entire solution process below:

#### Explanation:

We can rewrite the expression within the parenthesis without exponents as:

${\left(w \cdot w \cdot w \cdot x \cdot x \cdot x \cdot x \cdot y \cdot y \cdot y \cdot y \cdot y \cdot y\right)}^{2}$

We can now rewrite this as:

$\left(w \cdot w \cdot w \cdot x \cdot x \cdot x \cdot x \cdot y \cdot y \cdot y \cdot y \cdot y \cdot y\right) \left(w \cdot w \cdot w \cdot x \cdot x \cdot x \cdot x \cdot y \cdot y \cdot y \cdot y \cdot y \cdot y\right)$

Or

$\left(w \cdot w \cdot w \cdot w \cdot w \cdot w \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot y \cdot y \cdot y \cdot y \cdot y \cdot y \cdot y \cdot y \cdot y \cdot y \cdot y \cdot y\right)$

Or

$\left(w \cdot w \cdot w \cdot w \cdot w \cdot w\right) \left(x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x\right) \left(y \cdot y \cdot y \cdot y \cdot y \cdot y \cdot y \cdot y \cdot y \cdot y \cdot y \cdot y\right)$

We can order the list of $w , x , y$ terms for multiplication in
$29 , 002 , 073 , 100$ ways.

#### Explanation:

I'm going to take a bit of a different take on this question.

We start with ${\left({w}^{3} {x}^{4} {y}^{6}\right)}^{2}$ and distribute the square:

${w}^{3 \times 2} {x}^{4 \times 2} {y}^{6 \times 2} = {w}^{6} {x}^{8} {y}^{12}$

This means there are a total of 6 $w$ terms, 8 $x$ terms, and 12 $y$ terms all multiplying each other. Since we multiplication in this sense has the commutative power (in that $a b = b a$) and so the letters can be written in any order, I can write:

• $w x y w x y \ldots$
• $x y y w x y \ldots$

or any other order, just so long as I list all the variables together as being multiplied (I'm ignoring ordering via brackets)

And so there are 26 variables in total with a group of 6, a group of 8, and a group of 12. This means we can take 26! as the numerator of a fraction which will express the ways 26 terms can be arranged, and then divide by 6!8!12! to eliminate counting of duplicate ordering:

(26!)/(6!8!12!)=>

(26xx25xxcancelcolor(blue)(24)xx23xxcancelcolor(violet)22^11xxcancelcolor(orange)(21)xxcancelcolor(pink)(20)xx19xxcancelcolor(purple)(18)^3xx17xxcancelcolor(brown)(16)xxcancelcolor(green)(15)xx14xx13xxcancelcolor(red)(12!))/(cancelcolor(blue)(6)xxcancelcolor(green)(5)xxcancelcolor(blue)(4)xxcancelcolor(green)(3)xxcancelcolor(brown)(2xx8)xxcancelcolor(orange)(7)xxcancelcolor(purple)(6)xxcancelcolor(pink)(5xx4)xxcancelcolor(orange)(3)xxcancelcolor(violet)(2)xxcancelcolor(red)(12!))=>

$26 \times 25 \times 23 \times 11 \times 19 \times 3 \times 17 \times 14 \times 13 \implies$

$29 , 002 , 073 , 100$