Making $61$ by inserting arithmetical operations between $5\;5\;7\;3$. (The Australian Carriage Game (Number Puzzle))

Question

I know this isn't really the theme of what is going on here, but this problem is killing me, so I thought you guys could help.

The Australian Carriage Game comes from the fact that Australian trains have 2 letters and 4 numbers making up a carriage code. The rules are using these 4 numbers in their exact order, you must make the given number (In this case, 61) by inserting mathematical signs and symbols in between each character. An example of such a solution would be: (Keep in mind you normally aim for 10, but I was challenged to make 61)

$$3 3 3 6 \quad\to\quad 3/3 + 3 + 6 = 10$$

A list below shows the acceptable functions:

1. Addition
2. Subtraction
3. Division
4. Multiplication
5. Factorials
6. Bracket use
7. Exponentials and roots ONLY if the numbers are involved (i.e. you can't say $5^0$ unless $5$ and $0$ were next to each other in that order)
8. Permutations and combinations (eg, if $6$ and $5$ were next to each other, in that order, you could use $6 \choose 5$).
9. Concatenation (eg, if $6$ and $7$ were next to each other, you can combine them to make $67$)

Unacceptable functions include

1. Floor and ceiling functions
2. Use of $e$
3. Double and triple (and so on) factorials, and derangements (subfactorials)

If you want to present any other functions that could be used, ask me and I'll confirm.

The specific problem is to make $61$ from only the numbers $5\;5\;7\;3$.

Remember: All four numbers must be used, in the order given.

Note: I have no reason to believe that there is actually a solution.

I can make $60$ (as $5(5 + 7!/((3!)!))$) and $62$ (in some simple way I forgot, sorry), but I can't get to $61$.

Can anyone help me out?

EDIT: Credit for Michael Albanese's solution of $5!! + 5!! + 7!!! + 3$, using the double and triple "multifactorial" (eg, $5!!=5\cdot (5-2)\cdot (5-4)=15$ and $7!!!=7\cdot (7-3)\cdot (7-6)=28$). I don't consider this a viable solution and the question is still open, but it was a route I hadn't considered, and definitely works!

Thanks!

Answer 1

If subfactorial !n is allowed (http://mathworld.wolfram.com/Subfactorial.html), you can do it via

$ 55+7-(!(!3))=61 $