Different combinations without replacement and indistinguishable objects I'm a bit of a math novice, but have been trying to come up with an answer to a question involving my favorite game: Magic the Gathering.
I am wondering how many different combinations of 7 cards can I draw from a deck of 60 cards comprised of 4 indistinguishable copies of 15 different cards.
I've been able to use number combinations to get 60 pick 7 being 386,206,920 but thought that assumed that each card was unique.
Thank You,
WJ
 A: Let $x_k$ be the number of cards of type $k$.  Then
$$x_1 + x_2 + x_3 + \ldots + x_{13} + x_{14} + x_{15} = 7 \tag{1}$$
is an equation in the nonnegative integers.  A particular solution of equation 1 corresponds to the placement of $14$ addition signs in a row of $7$ ones.  For instance,
$$+ + 1 1 + + 1 1 1 + + + + + 1 + + + + + 1$$
corresponds to the solution $x_1 = x_2 = 0$, $x_3 = 2$, $x_4 = 0$, $x_5 = 3$, $x_6 = x_7 = x_8 = x_9 = 0$, $x_{10} = 1$, $x_{11} = x_{12} = x_{13} = x_{14} = 0$, $x_{15} = 1$.  The number of solutions of equation 1 is
$$\binom{7 + 14}{14} = \binom{21}{14}$$
since we must choose which $14$ of the $21$ positions required for $7$ ones and $14$ addition signs will be filled with addition signs.
However, we have not addressed the requirement that $x_k \leq 4$ for $1 \leq k \leq 15$.  We must subtract the number of solutions that violate this condition from the total.
Suppose $x_1 > 4$.  Then $x_1' = x_1 - 5$ is a nonnegative integer.  Substituting $x_1' + 5$ for $x_1$ in equation 1 yields
\begin{align*}
x_1' + 5 + x_2 + x_3 + \ldots + x_{13} + x_{14} + x_{15} & = 7\\
x_1' + x_2 + x_3 + \ldots + x_{13} + x_{14} + x_{15} & = 2 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers.  Since a particular solution of equation 2 corresponds to the placement of $14$ addition signs in a row of $2$ ones, equation 2 has
$$\binom{14 + 2}{14} = \binom{16}{14}$$
solutions in the nonnegative integers.
By symmetry, there are
$$\binom{16}{14}$$
solutions of equation that violate the restriction $x_k \leq 4$ for each $k$, $1 \leq k \leq 15$.  Hence,
$$\binom{15}{1}\binom{16}{14}$$
of the solutions of equation 1 violate one of the restrictions.  It is not possible to violate two of the restrictions simultaneously since $2 \cdot 5 = 10 > 7$.
Thus, the number of ways of drawing seven cards from the deck is
$$\binom{21}{14} - \binom{15}{1}\binom{16}{14} = 114480$$
