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There is a deck with 60 cards, with 11 different types of cards. It contains 20 A cards, 4 B cards, 4 C cards, 4 D cards, 4 E cards, 4 F card, 4 G cards, 4 H cards, 4 I cards, 4 J cards and 4 K cards ( for a total of 60 cards).

We pick 7 cards to our starting hand. How many different different starting hands are there? The order of the cards in the hand does not matter.

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It's the number of solutions of $$a_0+a_1+\cdots+a_{10}=7,\quad0\le a_0\le7,\quad0\le a_i\le4{\rm\ for\ }1\le i\le10$$ This in turn is the coefficient of $x^7$ in $$(1+x+\cdots+x^7)(1+x+x^2+x^3+x^4)^{10}$$ which is $$(1-x^8)(1-x^5)^{10}(1-x)^{-11}$$ Ignoring terms $x^8$ and higher, this is $$(1-10x^5)(1+11x+66x^2+\cdots)$$ where the coefficient of $x^n$ in the second bracket is $10+n\choose n$. So it seems the answer is $${17\choose7}-660$$

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