The difference is that you are counting every possible permutation rather than every possible combination. For instance, the sequence $10,20,30,40,50$ is equivalent to $50,40,30,20,10.$ Since you have $5!$ equivalent permutation for each combination (given there are 5 numbers picked out of the big lot), you need to divide your result by $120$: giving the published odds of $1:175,711,536$.
As for your followup question, I just checked the game's rules and you win that prize iff you have the bonus number and none of the primaries.
The general equation you are looking for is the number of permutations of m balls picked out of n:
Getting none of the primary balls means getting a combination of all the numbers except those you picked (51!/46!) divided by all the possible combinations (56!/51!). So the odds of having none of the primary numbers is (very close to) 61% - the odds of that and having the bonus ball give a bit over 1:75 odds of winning that $2