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

2 Answers
Apr 10, 2017

Answer:

See the entire solution process below:

Explanation:

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

#(w*w*w*x*x*x*x*y*y*y*y*y*y)^2#

We can now rewrite this as:

#(w*w*w*x*x*x*x*y*y*y*y*y*y)(w*w*w*x*x*x*x*y*y*y*y*y*y)#

Or

#(w*w*w*w*w*w*x*x*x*x*x*x*x*x*y*y*y*y*y*y*y*y*y*y*y*y)#

Or

#(w*w*w*w*w*w)(x*x*x*x*x*x*x*x)(y*y*y*y*y*y*y*y*y*y*y*y)#

Answer:

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 #(w^3x^4y^6)^2# and distribute the square:

#w^(3xx2)x^(4xx2)y^(6xx2)=w^6x^8y^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 #ab=ba#) and so the letters can be written in any order, I can write:

  • #wxywxy...#
  • #xyywxy...#

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!))=>#

#26xx25xx23xx11xx19xx3xx17xx14xx13=>#

#29,002,073,100#