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Please someone help with this question.
Solve,
C(n,8)= C(n,4)
A**a**f
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Hi abbeycity,
Welcome to the forum!
Hint for Solve, C(n,8)= C(n,4).
C(n,3)=n!/(3!(n-3)!)
.
Similarly, C(n,4)=n!/(4!(n-4)!)
It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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hi abbeycity
Welcome to the forum.
I'll assume, like Ganesh, that C here means Combinations.
(n-4)! = (n-4)(n-5)(n-6)(n-7) (n-8)!.........
and
8! = 8 x 7 x 6 x 5 x 4!
So there are several simplification steps possible here. The n! can be cancelled, and the common brackets on each side can be cancelled. There's no reason why we shouldn't also invert the fractions to eliminate fractions completely.
As we know n is a whole number and both sides have 4 terms, in descending order of size we can conclude that the biggest terms must be equal, the next equal, and so on.
8 = n-4
7 = n-5
6 = n-6
5 = n-7
These all lead to the same answer. I'll leave you to decide what it is.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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