In trying to count the number of 13-card hands where there is at least one ace and no J, Q, K, we can see one way is $$ \sum_{k=1}^4 \binom{4}{k}\binom{36}{13-k} = 9722433280. $$
However, I cannot see why $$ \binom{4}{1}\binom{39}{12} $$ fails. I take one ace and add it to the hand, and put the remaining aces into the pile from which to choose. This will give me all hands with at least one ace.
$\binom{39}{12}$ is the number of hands with a fixed ace (say, the ace of hearts). But $\binom{39}{12}$ is also the number of hands with the ace of spades.
So: what is the number of hands having either the ace of spades or the ace of hearts? Is it $2\binom{39}{12}$? Well, of course not, since we are counting twice every hand having both the aces.
The same argument applies when you ask how many hands have either the ace of spades, or the ace of hearts, or the ace of clubs or the ace of diamonds. The answer is not $4\binom{39}{12}$ because you are counting twice (or more than twice) some hand.