Permutation/Combination Problem: solve for n in n+2C4 = 6(nC2)

[creamcheesedigiorno]

Last question on my assignment, and I really don't know where to start with it.

Solve for n in the equation: n+2C4 = 6(nC2)

I know from imputing numbers into the equation that the answer is 7, but I'm sure there's a more formulaic way to solve it. Would anybody be willing to help explain how I could solve this problem in a better way? Any help is greatly appreciated!

 

[Deleted member 4993]

Last question on my assignment, and I really don't know where to start with it.

Solve for n in the equation: n+2C4 = 6(nC2)

I know from imputing numbers into the equation that the answer is 7, but I'm sure there's a more formulaic way to solve it. Would anybody be willing to help explain how I could solve this problem in a better way? Any help is greatly appreciated!

I assume - you are writing:

_(n+2)C₄ = 6 * _nC₂

Then, write the expressions for

_(n+2)C₄ = ??

_nC₂ = ??

 

[creamcheesedigiorno]

I assume - you are writing:

_(n+2)C₄ = 6 * _nC₂

Then, write the expressions for

_(n+2)C₄ = ??

_nC₂ = ??

Sorry, I'm not following... I haven't practiced any math for quite some time so I'm pretty rusty. You're right about the equation, that's what it's meant to be. Thanks for the help.

 

[pka]

Last question on my assignment, and I really don't know where to start with it.
Solve for n in the equation: n+2C4 = 6(nC2)
the answer is 7

If you use then code

[ tex]\dbinom{n+2}{4}[ /tex] (without the last spaces) you get the binomial notation \(\displaystyle \dbinom{n+2}{4}\)

\(\displaystyle \dbinom{n+2}{4}=\dfrac{(n+2)(n+1)(n)(n-1)}{(4)(3)(2)(1)}~\&~6\dbinom{n}{2}=6\dfrac{(n)(n-1)}{(2)(1)}\)

 

[creamcheesedigiorno]

If you use then code

\(\displaystyle \dbinom{n+2}{4}=\dfrac{(n+2)(n+1)(n)(n-1)}{(4)(3)(2)(1)}~\&~6\dbinom{n}{2}=6\dfrac{(n)(n-1)}{(2)(1)}\)

Oh, I get it now, I didn't think of it that way before. So now I have (n+2)(n+1) = 72. Just one more question... I can see just by looking at it that the answer is 7. But is that how I'm actually supposed to get the answer? Or is there another step I could be doing after (n+2)(n+1) = 72 to get the final answer? Thanks again everyone!

 

[stapel]

Oh, I get it now, I didn't think of it that way before. So now I have (n+2)(n+1) = 72. Just one more question... I can see just by looking at it that the answer is 7. But is that how I'm actually supposed to get the answer? Or is there another step I could be doing after (n+2)(n+1) = 72 to get the final answer? Thanks again everyone!

Okay; you've been given the meaning, the formula, and the equation to set up. You've posted this to "Intermediate/Advance Algebra", so you've already taken beginning algebra and learning how to solve equations. Try applying some of that information.

Please show some work of your own, so we can see where you're getting stuck. Thank you!

 

[creamcheesedigiorno]

Okay; you've been given the meaning, the formula, and the equation to set up. You've posted this to "Intermediate/Advance Algebra", so you've already taken beginning algebra and learning how to solve equations. Try applying some of that information.

Please show some work of your own, so we can see where you're getting stuck. Thank you!

Basically, I don't know where to go from with (n+2)(n+1) = 72. The only idea I have is to convert it to n² + 3n = 70, but then I'm stuck again. I guess I really just need to do a lot of algebra refreshing. I never learned it properly back when I was in school, so I've just been trying to pick it up as a go. I'm definitely going to have to go back an review the basics I guess!

 

[Deleted member 4993]

Basically, I don't know where to go from with (n+2)(n+1) = 72. The only idea I have is to convert it to n² + 3n = 70, (Correct) but then I'm stuck again. I guess I really just need to do a lot of algebra refreshing. I never learned it properly back when I was in school, so I've just been trying to pick it up as a go. I'm definitely going to have to go back an review the basics I guess!

Now you have a quadratic equation:

n² + 3n = 70

of the form:

An² + Bn + C = 0

What do you remember about solving "quadratic equations" (or roots of polynomials)?

Do a Google search!

 

[creamcheesedigiorno]

Now you have a quadratic equation:

n² + 3n = 70

of the form:

An² + Bn + C = 0

What do you remember about solving "quadratic equations" (or roots of polynomials)?

Do a Google search!

I remember... nothing! Lol. Alright, will do. I had been trying google but wasn't finding what I was looking for. But I will now, thanks for pointing me on the right track!

 

[stapel]

I remember... nothing!

Didn't your class just cover this material, before assigning this homework problem? :shock:

 

[creamcheesedigiorno]

Didn't your class just cover this material, before assigning this homework problem? :shock:

I'm sure it was covered in my 11th grade pre-calc class - 5 years ago. Now I'm doing my grade 12 through distance learning. I'm doing fine following the new concepts, it's just that my foundation is poor.

 

[stapel]

I'm sure it was covered in my 11th grade pre-calc class - 5 years ago. Now I'm doing my grade 12 through distance learning. I'm doing fine following the new concepts, it's just that my foundation is poor.

Ah. They were supposed to have tested you first, so they'd know exactly how much you remembered (and then put you in a class for which you were prepared), precisely so you wouldn't be placed in this situation. :shock: