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
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 $$\binom{21}{14} - \binom{15}{1}\binom{16}{14} = 114480$$