# Balls arrangements in a line

We randomly arrange balls numbered 1-100 in a line. What is the probability that there is a spot which splits the balls into two groups: All the balls preceding the splitting point are placed in ascending order (regarding their number), and all the balls from that point forward are place in decending order?

What I did: 100 different balls are placed in a line. Therefore $|\Omega| = 100!$

There are 100 possible "split-points". We'll mark $N$="split-point". For every $N$, the possibility of it being a split-point is $\binom{100}{N}$, as we choose the first $N$ balls and have only one way to arrange them and the remaining $100-N$ balls. Therefore: $$P(\text{a split point exists})=\frac{\sum\limits_{N=1}^{100}\binom{100}{N}}{100!}=\frac{\sum\limits_{N=0}^{100}\binom{100}{N}-1}{100!}=\frac{2^{100}-1}{100!}$$

But the answer given is:$$\frac{2^{99}}{100!}$$

Am I wrong? How?

Thank you for your time and effort.

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