SOLUTION: Dear math teacher, I am having difficulties solving for n in the following problem: nC12 = nC8 n!/(n-12)!12! = n!/(n-8)!8! n(n-1)...(n-11)(n-12)!/(n-12)!12! = n(n-1)...(n-

Question 476326: Dear math teacher,
I am having difficulties solving for n in the following problem:
nC12 = nC8
n!/(n-12)!12! = n!/(n-8)!8!
n(n-1)...(n-11)(n-12)!/(n-12)!12! = n(n-1)...(n-7)(n-8)!/(n-8)!8!
(n-8)(n-9)(n-10)(n-11)/12! = 1/8!
n^4-21n^3+110n^2-17n^3+357n^2-1870n+72n^2-1512n+7920 / 12! = 1/8!
Would you please let me know whether I am on the right track? My solving steps seem convoluted like a kidney tubule! Please let me know where I made an error.
Thank you very much.
Yours,
I.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
you're on the right track, but you needed to do some other type thinking before getting to the solution, which is a rule that i will explain to you after.
the equation is:
nC12 = nC8
this formula translates to:
n!/(12!*(n-12)!) = n!/(8!*(n-8)!)
cross multiply to get:
n! * (8!*(n-8)!) = n! * (12!*(n-12)!)
divide both sides of this equation by n! to get:
8! * (n-8)! = 12! * (n-12)!
divide both sides of this equation by 8! and divide both sides of this equation by (n-8)! to get:
(n-8)! / (n-12)! = 12!/8!
12! is equal to 12 * 11 * 10 * 9
(n-8)! / (-12)! is equal to (n-8)*(n-9)*(n-10)*(n-11)*(n-12)! / (n-12)! which is equal to:
(n-8)*(n-9)*(n-10)*(n-11)
our equation becomes:
(n-8)*(n-9)*(n-10)*(n-11) = 12*11*10*9
the left side of this equation will be equal to the right side of this equation if:
n-8 = 12
n-9 = 11
n-10 = 10
n-11 = 9
all of these equation would be true if n = 20.
that's your answer.
n = 20.
20C12 = 125970
20C8 = 125970
they're the same.
note that 12 + 8 = 20
that's the rule.
you can use it with any problems of this type to immediately get the answer.
an example:
nC13 = nC22
13 + 22 = 35
your answer is n = 35
35C13 = 1476337800
35C22 = 1476337800
The basic properties of combinations allows this to happen.
that property is that nCx = nC(n-x)
from that formula, you get x + (n-x) = n
a simpler example:
take n = 7
your values for each of the possible combinations are:
7C0 = 1
7C1 = 7
7C2 = 21
7C3 = 35
7C4 = 35
7C5 = 21
7C6 = 7
7C7 = 1
you can see that 7Cx = 7C(7-x)
this makes 7C0 = 7C7
this makes 7C1 = 7C6
this makes 7C2 = 7C5
etc.
Add the x and the (n-x) together and you get 7 every time.
your problem as nC12 = nC8
add 12 and 8 together and you get 20.

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