Find the least number of comparisons needed to sort (order) five elements and
devise an algorithm that sorts these elements using this number of comparisons.

> 
> *[Solution](http://www.math.mcgill.ca/olga/34006midsol.pdf):* There are 5! = 120 possible outcomes. Therefore a binary
>  tree for the sorting procedure will have at least 7 levels. Indeed, $2^h$
>  ≥ 120 implies $h $ ≥ 7. But 7 comparisons is not enough. The least number
>  of comparisons needed to sort (order) five elements is 8.
> 
> 
> 

Here is my actual question: I did find an algorithm that does it in 8 comparison but how can I prove that it *can't* be done in 7 comparisons?