Proof of solutions to “24” operations game Consider a deck of cards with values $1$ through $13$, each with multiplicity $4$, so that $$S = \left( \bigcup_{i=1}^{13} \{ i \} \right) \times \{1,2,3,4\}$$
Supposedly, there exists a game in which four cards $(v,m)$ in $S$ are selected (here $m$ stands in for the suit), and participants try to devise a series of operations using:


*

*addition $+$;

*subtraction $-$;

*multiplication $\times$;

*division $\div$;

*exponentiation $\wedge$;

*factorial-ization $!$; and

*any number of parentheses $()$


on the $v$-values that yields $24$. For example,
$$(2, 1) \times (10,3) - (2,4) + (3,2)! = 24$$



*

*Do there exist any combinations of cards for which no series of operations yielding $24$ exists?

*What mathematical processes could one use to analyze other aspects of solutions, such as how many solutions exist?


This is a post-examination game we are playing in class to kill time. I haven’t the slightest idea of how one would approach any sort of ‘proof.’
 A: Dark Malthorp showed that there are combinations of cards that cannot yield 24 from those operations you specified. I'll try to answer your other question about what strategy could someone use. Given the number/types of allowed operations, I think this strategy is unfeasible but it's only to showcase what we can do if the game was less complex (no ^, smaller numbers, etc.). EDIT: See comments on why $!$ cannot be removed.
Every formula built from $v_1,v_2,v_3,$ and $v_4$, using the operations $+,-,\cdot,/,$^$,$ and $!$, is of the following five forms:


*

*$(((v_1!_n*v_2!_n)!_n*v_3!_n)!_n*v_4!_n)!_n$

*$((v_1!_n*(v_2!_n*v_3!_n)!_n)!_n*v_4!_n)!_n$

*$(v_1!_n*((v_2!_n*v_3!_n)!_n*v_4!_n)!_n)!_n$

*$(v_1!_n*(v_2!_n*(v_3!_n*v_4!_n)!_n)!_n)!_n$

*$((v_1!_n*v_2!_n)!_n*(v_3!_n*v_4!_n)!_n)!_n$


where $*$ is a binary operation $+,-,\cdot,/,$ or ^, and $!_n$ is an $n$ number of factorials $!$. It looks really ugly since a $!_n$ may appear after every $v_n$ and parenthesis.
I had a feeling that $v_1=v_2=v_3=v_4=13$ would work. Informally, making all the numbers the same "limits" what the operations can do in a sense, and Dark confirmed that 13,13,13,13 is one of the combinations that cannot yield 24.
So let's try proof by contradiction. This is in no way a proof since I haven't actually checked every case.

Suppose $v_1=v_2=v_3=v_4=13$, and that every form yields 24. Then in every form, $n$ in the outermost $!_n$ is zero.* (* means not fully justified without a lot of case checking). First, we have that $m!=24$ iff $m=4$. Suppose $n=2$, then the same formula with outermost $!_{n-1}$ in place of $!_n$ must yield $4$ (since then $4!=24$). In this case we require $\phi!_1$=4, where $\phi$ stands for the formula without the outermost $!_{n-1}=!_1$. But there is no number $m$ s.t. $m!=4$, contradiction, so $n\neq2$. We repeat this argument for $n>2$. Okay now is the part where we use $v_1=v_2=v_3=v_4=13$, and where I don't check the cases. Suppose $n=1$, then we require $\phi!_0=\phi=4$, which I do not believe is possible, so $n=0$.* (To actually check this would require a lot of computing). Then $\ldots$

As you can see, this devolves into A LOT of case checking. We'd have to eventually check the $!_n$'s in $\phi$ and subsequent subformulas are $!_0$.* I believe this as informally, getting $13!$ back to a lower number requires division by at least $13!$, taking up an operation slot. Maybe this is untrue of smaller numbers; this is another reason I chose 13,13,13,13 and not some other quadruple of same numbers, along with the fact that it doesn't seem you can get factors of 24 easily from 13.
Anyways by fixing ALL the $!_n$'s with a finite number of $n$'s, and noting there are only a finite number of combinations of binary operations, you'd have a finite number of cases to check.
I hope this helps in your future problems.
