Jtdylktuy

Can someone please help me simplify the below problem. I am trying to help my niece

$${24C^4 over 4C^{12}}$$

I believe that the simplified answer is $6/C^8$ because $4$ goes into $24$ $6$ times and $6$ remains as the numerator. Also, you need to subtract $4$ from $C^4$ which becomes $1$ and subtract $4$ from $C^{12}$ and it equals $C^8$. however my niece thinks it is: $1/(6C^8)$. what is the correct answer?

One approach I've used for someone beginning to work with this type of situation is the following, although I'd probably replace it with something similar and simpler, such as $frac{24C^2}{4C^5},$ to actually illustrate the idea.

$$ frac{24C^4}{4C^{12}} ; = ; frac{24 cdot C^4}{4 cdot C^{12}} ; = ; frac{24}{4} cdot frac{C^4}{C^{12}} $$

$$ = ; ; frac{24}{4} cdot frac{C cdot C cdot C cdot C cdot 1 cdot 1cdot 1cdot 1cdot 1cdot 1cdot 1cdot 1}{C cdot C cdot C cdot C cdot C cdot C cdot C cdot C cdot C cdot C cdot C cdot C} $$

$$ = ;; frac{24}{4} cdot frac{C}{C} cdot frac{C}{C} cdot frac{C}{C} cdot frac{C}{C} cdot frac{1}{C} cdot frac{1}{C} cdot frac{1}{C} cdot frac{1}{C} cdot frac{1}{C} cdot frac{1}{C} cdot frac{1}{C} cdot frac{1}{C} $$

$$ = ;; frac{6}{1} cdot 1 cdot 1 cdot 1 cdot 1 cdot frac{1}{C} cdot frac{1}{C} cdot frac{1}{C} cdot frac{1}{C} cdot frac{1}{C} cdot frac{1}{C} cdot frac{1}{C} cdot frac{1}{C} $$

$$ = ;; frac{6}{1} cdot frac{1 cdot 1 cdot 1 cdot 1 cdot 1 cdot 1 cdot 1 cdot 1 }{C cdot C cdot C cdot C cdot C cdot C cdot C cdot C}$$

$$ = ;; frac{6}{1} cdot frac{1}{C^8} ;; = ;; frac{6}{C^8} $$

What you want to do is go over several elaborate expansions like this and let the student discover the shortcuts that can be taken, and then when the student is comfortable with simple things like this (integer times a letter to a positive integer power, divided by an integer times a letter to a positive integer power), begin branching out to things like $frac{24x^3y^5}{6x^2y^8}.$ All this should not take more than 5 to 15 minutes, after which you can begin talking about the relevance of the laws of exponents to the shortcuts the student has observed, as well as mixing in things like $C^3 cdot C^4 = (CCC)cdot (CCCC) = CCCCCCC = C^{3+4}$ $(3;C$'s followed by $4;C$'s is $7;C$'s), if this had not already been done at some previous time.

Yes you are correct. The simplified answer is $frac{6}{C^8}$.