Question 476308: Dear math teacher,
I am having difficulties with the following problem:
4 times nC2 = (n+2)C3, (a)solve for n; (b) n cannot equal to or must be different from ...integers.
4*n(n-1)*(n-2)!/(n-2)!(2.1) = (n+2)!/(n+2-3)!*(3.2.1) and after canceling out 2(n-2)! we get
2n(n-1) = (n+2)(n+1)n(n-1)!/(n-1)!(3.2.1) and after canceling out n and (n-1)! we get
2(n-1) = (n+2)(n+1)/6
n^2 - 9n + 14 = 0
(n-7)(n-2) = 0
n=7 and n=2
To answer part (a) of the problem, I say the solution is 7 only since (n-2) was canceled out above. For part (b) n cannot equal to 0, 1, and 2 since n canceled out, (n-1) canceled out, and (n-2) term canceled out.
Please let me know, if I reason through this correctly. Thank you very much.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! you calculated the answer correctly.
n is equal to 7 or n is equal to 2.
that takes care of part a.
both of those values of n will make the equations equal.
that was confirmed.
part b says n cannot be equal to, or must be different from ... integers?
not sure i understand that, but, if i must pick a number that should be disallowed, it would be n = 2.
your equation is:
4*nC2 = (n+2)C3
this equation can be translated to:
(4*n!)/(2!(n-2)!) = (n+2)!/(3!(n+2-3)!) which becomes:
(4*n!)/(2!(n-2)!) = (n+2)!/(3!(n-1)!)
when n = 7, this equation becomes:
4*7!/(2!*5!) = 9!/(3!*6!)
when n = 2, this equation becomes:
4*2!/(2!*0!) = 4!/(3!*1!)
the left side of this equation have n equal to 2 which is equal to the number of positions in the subgroup you are trying to create.
if that's what they're referring to, then n = 2 would be disallowed because the subgroup is the same size as the group.
outside of that I'm not exactly sure what their complaint is.
assuming i understood correctly, i would say that n = 2 would be disallowed because the group created is the same size as the subgroup that would be created from it.
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