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Arrange the five numbers 1, 2, 3, 4, 5 in a circle, in some order. We can place the 5 at the top; let the other numbers be $a,$ $b,$ $c,$ $d.$ Then the sum we are interested in is the sum of the product of adjacent pairs.



Assume that the numbers have been arranged so that the sum we are interested in has been maximized. The sum for this arrangement is $5a+ab+bc+cd+5d.$ This means that if we were to change the arrangement, the sum must either stay the same or decrease.

Suppose we swap 5 and $a$:



The sum is now $5a+5b+bc+cd+ad.$ Hence, $$5a+5b+bc+cd+ad\le 5a+ab+bc+cd+5d.$$This reduces to $ab-ad+5d-5b\ge 0,$ which factors as $(5-a)(d-b)\ge 0.$ We know $5-a\ge 0,$ so $d-b\ge 0.$ And since $b$ and $d$ are distinct, $d>b.$

Now, suppose we swap 5 and $d$:



The sum is now $ad+ab+bc+5c+5d.$ Hence, $$ad+ab+bc+5c+5d\le 5a+ab+bc+cd+5d.$$This reduces to $cd-ad+5a-5c\ge 0,$ which factors as $(5-d)(a-c)\ge 0.$ We know $5-d\ge 0,$ so $a-c\ge 0.$ And since $a$ and $c$ are distinct, $a>c.$

Finally, by reflecting the diagram along the vertical axis, we can assume that $b>c.$ This leaves three cases to check: $$\begin{array}{ccccc}a& b& c& d& 5a+ab+bc+cd+5d\\ 2& 3& 1& 4& 43\\ 3& 2& 1& 4& 47\\ 4& 2& 1& 3& 48\end{array}$$Hence, the largest possible sum is 48. Furthermore, there are ten permutations that work: The five cyclic permutations of $(5,4,2,1,3),$ and the five cyclic permutations of its reverse, namely $(5,3,1,2,4).$ Thus, $M+N=48+10=\boxed{58}.$