EDIT: there was a bug in my code so it missed some solutions. There are actually only 48 unsolvable hands, not 56.
EDIT 2: due to rounding errors, my code thought [9,7,5,5] was solvable, when in fact it is not (see update below).
I have played that game too, so when I saw your question I was pretty sure there are some hands that have no series reaching 24. Since there's only $\binom{52}{4} = 270,725$ possible hands of 4 cards (even less since this game doesn't count suits), I figured it would be feasible to do a computer search to find all of these solutions (see the bottom for my code).
My program found solutions for all but 48 hands (ignoring different suits):
[1,1,7,7] [1,1,9,9] [1,1,10,10] [1,1,10,11] [1,7,7,13] [1,9,9,9] [1,9,10,10] [1,10,10,10] [1,10,10,11] [1,10,11,11] [1,11,11,11] [1,13,13,13] [6,6,6,13] [6,6,7,7] [6,6,7,13] [6,6,13,13] [6,7,7,13] [6,7,13,13] [6,8,8,13] [6,11,11,13] [7,7,7,7] [7,7,7,13] [7,7,10,10] [7,7,10,12] [7,7,11,11] [7,7,13,13] [7,8,9,9] [7,10,10,13] [7,11,11,13] [7,13,13,13] [8,8,9,9] [8,8,11,11] [8,9,9,10] [8,9,13,13] [9,9,9,9] [9,9,9,11] [9,9,10,10] [9,9,13,13] [9,10,10,10] [9,10,10,11] [9,10,11,11], [10,10,10,10] [10,10,10,11] [10,10,11,11] [10,10,13,13] [10,11,11,11] [11,11,11,11] [13,13,13,13]
Counting suits, this works out to be 2413 possibilities out of the 270,725 total hands, or a probability of just a tad below 0.9%.
Since my program did not use factorials of any numbers larger than 13, so it is possible that some of these 48 actually do have solutions, but I don't find that very likely. Among these combinations all have repeated cards. None of the combinations have a 2, 3, 4, or 5, and only one has a 12. Surprisingly, the most common number in the unsolvable hands is 10 (accounting for suits, it shows up in 1109 out of the 2413 unsolvable hands). As someone who has played this game a lot, I was expecting 11 or 13 to be the most likely to give an unsolvable hand, but they only showed up in 845 and 1073 unsolvable hands, respectively.
As for the number of ways to get to 24 from a given hand, my program didn't really look at that because I was trying not to make the program take up to much computational power, and it was much easier to simply remember a single boolean than a whole chain.
Rounding errors update:
[9,7,5,5] is not a solution, but my code thinks it is due to rounding errors. For example, it couldn't distinguish $9^{5-7!}$ from $0$, and found this "solution":$$
\left(5 - \left(9^{5-7!}\right)!\right)! \approx 24
$$
If you use the gamma function to interpolate the factorials, this is off by only about$5\times 10^{-4804}$.
Removing exponentiation from the options, other than using $1^a = 1$, shows that this is the only solution that took advantage of this particular numerical error. Since I didn't ever take a factorial of a number bigger than 13, the factorials should not have introduced any other fake solutions.
Some interesting example hands:
Among the hands that have solutions, I've chosen to highlight a few particularly interesting ones:
[13,11,10,6], [13,6,1,1], [13,12,10,8], [11,11,5,5], [11,11,5,1], [9,7,7,7], and [5,1,1,1]
If you want to challenge yourself, see if you can solve these before looking at the solutions.
[13,11,10,6]
This is the only hand that requires a factorial of a number bigger than 10:
$$24 = \left(\frac{13 + \frac{11!}{10!}}{6}\right)!$$
[13,6,1,1]
This and [9,8,1,1] are the only ones that require use of exponentiation:
\begin{eqnarray} 24&=&\left(\frac8{1+1^9}\right)!\\24&=&\left(6 - 1^{13} - 1\right)! \end{eqnarray}
[9,7,7,7]
This hand has only one possible solution, and is an interesting example of one which requires factorials but doesn't use $4!=24$.
$$24 = \frac{9!}{7!+7!+7!}$$
If you disallow factorials, there are fully 430 unsolvable hands (not counting suits). Some of my favorite solutions without factorials include [11,11,1,5], [13,12,10,8], [11,11,5,5] and [5,1,1,1], which have the following unique non-factorial solutions:
[13,12,10,8]
$$24=\frac{10\cdot 12}{13-8}$$
[11,11,5,5]
$$24=5\cdot5-\frac{11}{11}$$
[11,11,5,1]
$$24=\frac{11\cdot 11 - 1}{5}$$
[5,1,1,1]
$$5^{1+1} - 1$$
Below is my Haskell program. Programming is not my strong point, so excuse the messiness.
unsolvable = p 4
-- all hands that have no way to reach 24.
-- derivation:
-- returns True if an input four card hand is solvable.
check :: (Num a, Fractional a, Eq a, Ord a, Floating a, Enum a) => [a] -> Bool
check [a,b,c,d] = firstCheck a b c d
check _ = False
--check as a four-argument function
firstCheck :: (Num a, Fractional a, Eq a, Ord a, Floating a, Enum a) => a -> a -> a -> a -> Bool
firstCheck a b c d = or [fCheck a b c d, fCheck b c d a, fCheck c d a b, fCheck d a b c, fCheck d b a c, fCheck a c b d]
fCheck :: (Num a, Fractional a, Eq a, Ord a, Floating a, Enum a) => a -> a -> a -> a -> Bool
fCheck a b c d = or $ map (secondCheck a b) $ op c d
{--finalCheck :: (Num a, Fractional a, Eq a, Ord a, Floating a, Enum a) => a -> a -> Bool
finalCheck a b = (24 `elem` s)
where s = op a b --}
--check if 3 numbers can make 24
sCheck :: (Num a, Fractional a, Eq a, Ord a, Floating a, Enum a) => a -> a -> a -> Bool
sCheck a b c = or $ map (finalCheck a) $ op b c
secondCheck :: (Num a, Fractional a, Eq a, Ord a, Floating a, Enum a) => a -> a -> a -> Bool
secondCheck a b c = (sCheck a b c) || (sCheck b a c) || (sCheck c a b)
-- checks if 2 numbers can combine to 24 or 4! = 24
finalCheck :: (Num a, Fractional a, Eq a, Ord a, Floating a, Enum a) => a -> a -> Bool
finalCheck a b = (24 `elem` s) || (4 `elem` s)
where s = op a b
--all operations except factorial, with the assumption that a a -> a -> [a]
opp 0 b = [0,1,b,-b, 1+b, b-1, 1-b]
opp 1 b = [1+b,b,1/b,1-b,b-1,1]
opp a b = [a+b,a*b,a/b,a-b,b-a,b/a,a**b,b**a]
-- op includes operations from opp as well as manually including factorials up to 13!
op :: (Num a, Fractional a, Eq a, Ord a, Floating a, Enum a) => a -> a -> [a]
op 3 b = opp (min 3 b) (max 3 b) ++ op b 6 ++ op 6 b
op 4 b = opp (min 4 b) (max 4 b) ++ op b 24
op 5 b = opp (min 5 b) (max 5 b) ++ op b 120
op 6 b = opp (min 6 b) (max 6 b) ++ op b 720
op 7 b = opp (min 7 b) (max 7 b) ++ op b 5040
op 8 b = opp (min 8 b) (max 8 b) ++ op b 40320
op 9 b = opp (min 9 b) (max 9 b) ++ op b 362880
op 10 b = opp (min 10 b) (max 10 b) ++ op b 3628800
op 11 b = opp (min 11 b) (max 11 b) ++ op b 39916800
op 12 b = opp (min 12 b) (max 12 b) ++ op b 479001600
op 13 b = opp (min 13 b) (max 13 b) ++ op b 6227020800
op a b = opp (min a b) (max a b)
lcheck :: (Num a, Fractional a, Eq a, Ord a, Floating a, Enum a,Enum a) => [a] -> [Bool]
lcheck [a,b,c,d] = [check [a,b,c,d]]
lcheck s = map (and . lcheck . (\n -> n:s)) [(head s)..13]
-- s must be in order from largest to smallest.
-- (lcheck s !! i) is True whenever there is a solvable hand sorted
-- from largest to smallest that ends with (head s + i):s
ccheck :: (Num a, Fractional a, Eq a, Ord a, Floating a, Enum a,Enum a) => [a] -> Bool
ccheck = and . lcheck
-- returns True if a sorted hand ending with s is solvable
p1 = filter (not . (\k -> (ccheck [k])) . fromIntegral) [1..13]
-- possible smallest cards for an unsolvable hand
pf :: (Integral a) => [a] -> [[a]]
pf x = map (\s -> s:x) $ filter (not . (\k -> (ccheck $ map fromIntegral (k:x))) . fromIntegral) [(head x)..13]
-- if x is n cards, pf x lists possibilities of length n+1 endings for sorted hands that end in x
p :: Int -> [[Integer]]
p 1 = map (\x -> [x]) p1
p n = foldr (++) [] $ map pf $ p (n-1)
-- lists possible last n cards of a sorted unsolvable hand.
