It seems that the solitaire card game 'klondike' presents some head scatching when trying to work out odds for winnable games etc, but I would image a 'simpler' question is 'how many different deal combinations are there in the 'Klondike' card game?
[Math] Deal combinations in solitaire game ‘Klondike’
card-gamescombinatorics
Related Solutions
The numbers you quote are for "Thoughtful Solitaire", i.e. Klondike Solitare where the positions of all 52 cards are known.
So in theory it might be possible to look at all $52!\approx 8 \times 10^{67}$ permutations of the cards and for each one (or for a eighth of them, taking account of the equivalence of suits) see whether it is possible to solve that case or not with any of the many combinations of choices by looking at every combination of choices. In practice neither of those two options are practical.
To deal with the excessive number of permutations, one approach would be to take a random sample and to use statistical techniques to provide steadily narrowing confidence intervals around the estimates as the sample gets bigger.
To deal with the excessive number of choices, you can apply heuristics which provide good methods for taking decisions without investigating every final result. Doing this trims the decision tree and so shortens the time needed to investigate different possibilities. But even then, the consequences of different decisions in the game can sometimes have such far reaching and complicated consequences that not all initial permutations can be found to be solvable or not within a reasonable time. Ignoring those which do not produce a result quickly enough leads to the wide reported range for the probability.
There can be $2$ to $7$ different card values, and $k$ card values can occur in $\binom{14-k}k$ different value combinations. The number of ways of choosing $10$ cards with exactly $k$ different card values can be found using inclusion–exclusion: There are $\binom{8k}{10}$ ways to choose $10$ cards with values in a given set of $k$ values, so there are
$$ \sum_{j=2}^k(-1)^{k-j}\binom kj\binom{8j}{10} $$
ways to choose $10$ cards with values forming exactly a given set of $k$ values. Thus the number of combinations is
$$ \sum_{k=2}^7\binom{14-k}k\sum_{j=2}^k(-1)^{k-j}\binom kj\binom{8j}{10}=153020720928 $$
(computation) out of a total of $\binom{104}{10}=26100986351440$, so the probability for this to occur is
$$ \frac{153020720928}{26100986351440}=\frac{9563795058}{1631311646965}\approx0.00586264 $$
in agreement with Jonathan's simulations.
Best Answer
Assuing that you're referring to the standard version of the game as described at Wikipedia, and that the order of the cards in the deck is part of what distinguishes deals, the answer is $52!$, the number of different permutations of $52$ cards. If you disregard the order of the cards in the deck and are only interested in the different ways of dealing the $28$ cards on the table, the answer is $\displaystyle\frac{52!}{(52-28)!}$: There are $52$ choices for the first card to deal, $51$ for the next one, ..., and $52-27$ for the last one.