Discrete Math HW Problem

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teris

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I am trying to complete a study guide for my college class, APM 263 (discrete mathematics). However, I cannot for the life of me figure out how to do/solve/prove this problem. I will try my best to type it out; I'm no good at typing out complex math problems. Thank you for your help and insight.


(n sub 1) + 6(n sub 2) + 6(n sub 3) = n^3



Note: For the first "(n sub 1)", it means the following: (n!)/[1!(n-1)!]
 
teris said:
I am trying to complete a study guide for my college class, APM 263 (discrete mathematics). However, I cannot for the life of me figure out how to do/solve/prove this problem. I will try my best to type it out; I'm no good at typing out complex math problems. Thank you for your help and insight.


(n sub 1) + 6(n sub 2) + 6(n sub 3) = n^3



Note: For the first "(n sub 1)", it means the following: (n!)/[1!(n-1)!]
Click to expand...

this is a simple algebra problem - just expand the given expressions. For example:

\(\displaystyle n_1 = \frac{n!}{1!\cdot (n-1)!} \, = \, \frac{n!}{(n-1)!} \, = \, \frac{n \cdot (n-1)!}{(n-1)!} = \, n\)

similarly find the other two terms and add those up.
 
So for the second term, does the following make sense?

6 [ (n!) / [ 2!(n-2)! ] = 6 (n!) / [ 2(n-2)!] = 3n(n-2)!/(n-2)! = 3

I think I'm getting really messed up in the middle. Thank you for your help. I really appreciate this.
 
teris said:
So for the second term, does the following make sense?

6 [ (n!) / [ 2!(n-2)! ] = 6 (n!) / [ 2(n-2)!] = 3n(n-2)!/(n-2)! = 3<---- Incorrect

\(\displaystyle n! = n\cdot (n-1) \cdot (n-2) \cdot (n-3) \cdot (n-4) \cdot (n-5) \cdot (n-6) \cdot (n-7)\cdot\cdot\cdot 3\cdot 2\cdot 1\)

I think I'm getting really messed up in the middle. Thank you for your help. I really appreciate this.
Click to expand...
 
thank you but the biggest part where i get hung up is how to get n^3 out of the LHS when the terms are going to be added. if i get the terms to cancel out into n+n+n=3n
 
Write out the terms - correctly - and add. Correct the second term - using the method I showed you.

Show us what you are doing.
 
ok the first term we get n, the second term looks like 6 n(n-1)(n-2)!/2!(n-2)! where i cancel out the (n-2)! and end up with 3n(n-1), on the third term i break down the n! into n x n-1 x n-2 x (n-3)!/(n-3)! where you cancel out the (n-3)!, so for the remaining n x n-1 x n-2 do you foil it?
 
teris said:
I apologize for my lack of experience with factorials
Click to expand...


You seem to be doing fine with factorials. You correctly reduced the ratio of factorial expressions in the second term to 3n(n - 1). You also correctly reduced the third given term to n(n - 1)(n - 2). I would say that those symbolic cancellations are the hardest part of this exercise.

I think your lack may be in confidence. If you're wondering whether or not to use FOIL at some point in any exercise, what keeps you from trying? Don't stop. Keep going.

Yes -- you need to multiply everything out in order to see how the like terms cancel. Use FOIL to expand (n - 1)(n - 2), followed by distributing the n.

If you multiply everything out, you'll end up with an unsimplified cubic polynomial on the LHS:

n + 3n^2 - 3n + n^3 - 3n^2 + 2n

Now you're one step away from proving that the given LHS equals the given RHS.

MY EDIT: removed ambiguity regarding goal
 
teris said:
ok the first term we get n, the second term looks like 6 n(n-1)(n-2)!/2!(n-2)! where i cancel out the (n-2)! and end up with 3n(n-1), on the third term i break down the n! into n x n-1 x n-2 x (n-3)!/(n-3)! where you cancel out the (n-3)!, so for the remaining n x n-1 x n-2 do you foil it?
Click to expand...


I would not distribute the factors. Instead, simplify all three terms like you have done here, bring it all over to one side and factor as much as possible. "FOIL"ing everything out might lead to a factoring nightmare later on.
 
Looking at the form of the answer required - in this case - FOILing everything and simplifying is the best route to go.
 
Subhotosh Khan said:
Looking at the form of the answer required - in this case - FOILing everything and simplifying is the best route to go.
Click to expand...

Okay, you win. I just hope he won't try the same thing on a quintic :wink:
 
What would be Latin equivalent term for a seventh order polynomial - Septic???!! :mrgreen:
 
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