Question 475525: Dear math teacher,
Would you please explain why n cannot equal -4 and 5 as a solution to the following equation?
nC(n-2) = 10
Here is how I solved it:
{n(n-1)}/2 = 10
(n)^2 - n - 20 = 0
(n-5)times(n+4) = 0
n-5 = 0 and n+4 = 0
n = 5 and n = -4
Now, I know that we cannot have a negative factorial, which is why I only selected n = 5 for the answer to this equation. However, if I were to think deeper and say n cannot equal -4 because that would make the expression (n-5)times(n+4) = 0. But this expression is supposed to be equal to zero. So, let's say instead of -4, I got 6, then the 6 would also be a solution to the equation and then, we would have n = {5, 6}.
My second question is: Can n = 2? I don't think so because if n = 2 then the expression {n(n-1)times(n-2)!}/{(2.1)times(n-2)!} = 10 would also become zero. Therefore, n cannot ever equal 2. Is that the correct reasoning?
My third question is: Can n = 0? LOGICALLY, IF I PLUG-IN ZERO IN {n(n-1)times(n-2)!}/{(2.1)times(n-2)!} = 10, THE ENTIRE EXPRESSION WILL BECOME ZERO. SO CAN n EVER EQUAL ZERO OR NO (OF COURSE HYPOTHETICALLY SPEAKING)?
My forth question is: Can n = 1? Here again, the expression {n(n-1)times(n-2)!}/{(2.1)times(n-2)!} = 10 would become zero.
For some reason my textbook, when solving another problem says n cannot equal to a value that you are canceling out. In our problem, it would be n cannot equal to 2 since I am canceling out (n-2)! from numerator and denominator of the expression {n(n-1)times(n-2)!}/{(2.1)times(n-2)!} = 10
Therefore, in my case for the expression
nC(n-2) = 10
{n(n-1)}/2 = 10
(n)^2 - n - 20 = 0
(n-5)times(n+4) = 0
n-5 = 0 and n+4 = 0
n = 5 and n = -4, and n cannot equal 2 and cannot equal -4. But what is the reason n cannot equal 2 and -4? Can n ever equal 0 and -1?
Thank you so much for reading this long question and thank you for finding the time to answer it.
Yours sincerely,
I.
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
You are working way too hard. You properly derived the quadratic equation and obtained the correct solutions. You discard the negative root because the factorial function is not defined for negative integers. That leaves you with 5 as the only answer to the problem.
No, the answer cannot be 0 because (0 - 2)! for the same reason that you excluded the -4, i.e. the factorial function is not defined for negative integers. The LHS of your equation does NOT go to zero, first because of the undefined part of the denominator and second because your numerator of 0! = 1 NOT 0. See the definition of Factorial:
Wikipedia Factorial
No, the answer cannot be n = 2 because that would give you a zero factor in the denominator.
Take the term "cancelling out" and remove it from your mathematical vocabulary. You are dividing identical terms so as to achieve an overall multiplier of the fraction of 1. What your book is trying to tell you is that you cannot have a zero factor in your denominator. If n = 2, then n - 2 = 0.
Hope that helps.
John
My calculator said it, I believe it, that settles it
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