Combinatorics: Number of possible 10-card hands from superdeck (10 times 52 cards) I have the following problem from book "Introduction to Probability", p.32

A certain casino uses 10 standard decks of cards mixed together into one big deck, which
  we will call a superdeck. Thus, the superdeck has 52 · 10 = 520 cards, with 10 copies
  of each card. How many different 10-card hands can be dealt from the superdeck? The
  order of the cards does not matter, nor does it matter which of the original 10 decks
  the cards came from. Express your answer as a binomial coefficient.
  Hint: Bose-Einstein.

My solution:
Because the number of cards of each type in the superdeck (10) is not less than the size of the hand (10), and thus not limiting, it's the same as sampling with replacement where the order does not matter, 
so the number of possible 10-card hands would be $\binom{52+10-1}{10}$.
Is my thinking correct?

Just to address the "Bose-Einstein" hint:  in quantum physics, a set of particles are said to obey Bose-Einstein statistics when they are physically indistinguishable (so that there's no way, even in principle, to tell them apart);  and multiple particles can be in the same state.  A "State" is a pretty broad concept in physics, but you can think of these states as being like energy levels in an atom.  The other related concept is Fermi-Dirac statistics, in which the particles are indistinguishable and you can only have one particle in each state.  The best-known particle that obeys Bose-Einstein statistics is the photon;  the best-known particles that obey Fermi-Dirac statistics are the electron, proton, and neutron.
The "superdeck" problem posed is then the same as the following problem:  A system of 10 particles (i.e. cards in the hand) obeying Bose-Einstein statistics are put into a system with 52 distinct one-particle states (i.e., types of cards in a standard deck.)  What is the number of distinguishable states of the collective system?  The answer can be found by imagining partitioning the ten particles among the 52 one-particle states, and saying that we only care about the number of particles in each one-particle state.  This then becomes a standard "balls and walls" problem, with the answer of ${61 \choose 10} = 90 177 170 226$ as you found.
