Your reasoning was no so evident for me at first, so I decided to dig a little and what I first found was a table with an entry that agrees with your interpretation, but then I wanted to know where that multichoose formula comes from, and after a while I found useful this explanation using the bars and stars approach.
In your case, for example, we would represent the $10$ cards of the hand as "stars" and then we would put $52-1$ "bars" between them to represent the idea of $52$ different types$^1$ of cards. The stars that are to the left of a bar are of the same type. We have then to be careful with the extreme case where all the $10$ stars are to the left of a bar, meaning that the all the $10$ cards are all of the same type. As you and @JMoravitz correctly pointed out, this is possible because the size of the hand is not greater than the multiplicity of each type in the superdeck.
Under this approach we could imagine that we have a total of $52+10-1$ bins to place the stars and the bars, and then ask in how many ways we could place the $10$ stars out of a total $52+10-1$ available bins. Now the answer is more evident to me, in
ways. Once you place the stars there is only one way to put the bars, since the order does not matter.
EDIT: I was wondering why Bose-Einstein would be a hint in the problem. Following the perspective proposed in this video, it seems that this problem is analogous to the Bose-Einstein statistics in that we could thought of the $10$ cards (stars) in the hand as indistinguishable particles and the $52^*$ types of cards in a deck as energy states where those particles could condensate (don't get me wrong, I am not a quantum physicist). In any case, it's not clear how that would be a hint to solve the problem...any hint?
$^*$ Thanks to Michael's answer I understood that the number of energy states is $52$ and not $52+10-1$, as I originally wrote in my edit...although it still seems a rather sophisticated hint for me.
$^1$ Here the type of a card is its rank and suit.