Suppose that the integers from $1$ to $7$ are randomly ordered from left to right. If the ordering consists of an initial increasing sequence followed by a decreasing sequence, then we say that we have one run up followed by one run down.

What is the probability that a random ordering of the $7$ integers results in one run up followed by one run down.

My attempt: I think I did this right, but I am looking for verification or other interesting methods to approach this.

I noted that $7$ must not be in position $1$ or $7$. When considering $7$ to be in position $2$, you have $6\choose1$ different possibilities for position $1$, and positions $3$-$7$ would only have one possibility (the remaining digits in decreasing order). Similarly, we $6\choose2$, $6\choose3$. $6\choose4$, and $6\choose5$ possibilities when $7$ is in the $3^{rd}$, $4^{th}$, $5^{th}$, and $6^{th}$ spot, respectively.

This gives $p$ = ${{6\choose1}+{6\choose2}+{6\choose3}+{6\choose4}+{6\choose5}\over 7!}$ $=$ $.0123$

  • $\begingroup$ I am not sure, but I think the solution is valid. $\endgroup$
    – Peter
    Sep 6, 2017 at 22:10

1 Answer 1


This is $\dfrac{2^6-2}{7!}$ and would have been $\dfrac{2^6}{7!}$ if you had allowed the $7$ at an end (i.e. a run up or down of minimal length)

The $2^6$ has an easy combinatorial explanation: each of the digits $1,2,3,4,5,6$ can come either before or after the $7$, and once those decisions are made the pattern is fixed

  • $\begingroup$ I did much more work than I needed to. Thanks! $\endgroup$
    – Remy
    Sep 6, 2017 at 22:40

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