3

Can this be simplified to isolate the variable by itself?

$$\begin{align} \dfrac{\displaystyle{83 - x}\choose{\displaystyle x}}{\displaystyle{82}\choose{\displaystyle x}} \geq 0.50 \end{align}$$

This is where I am.

$$\begin{align} \dfrac{(83 - x)!}{(83 - 2x)!} \geq 0.50 \dfrac{82!}{(82 - x)!} \end{align}$$

Paul
  • 31
  • 1
    Why do you ask? (Maybe there is another way to solve the problem where this come from). – chubakueno Jan 20 '14 at 04:59
  • $$\prod_{k=0}^{x} \frac{83-x-k}{82-k} = \prod_{k=0}^x \big( 1+\frac{1-x}{82-k}\big) \ge 0.5$$ This may let you get a quick estimate to use a good starting value for an algorithm. – Jeff Snider Jan 20 '14 at 05:13

1 Answers1

0

I suppose you are looking for the lowest integer value of "x" which satisfies the inequality.

In my mind, an easy way would consist in using Jeff Snider's formulation and take the logarithms of both sides. So, the equation is just a sum of "x" logarithms.

If we assume that "x" is small, we could use the approximation Log(1+y) = y and if we also assume that "k" is small compared to 82, just forget the "k" in the denominator. All of this lead to the search of "x" such that

x (1-x) / 82 > Log(1/2)

Solve the quadratic equation assuming, for the time being that "x" is real : the only positive root is 8.06. Then try x=8; you get a value equal to 0.473355. Since your ratio decreases when "x" increases, try x=7; you get a value equal to 0.575047. So x=7 is your solution.

Just for testing puroposes, I replace the value of 0.5 by 0.01. Using the same method as before, the root of the quadratic equation is 19.94 while the solution is 17.

Claude Leibovici
  • 260,315