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As in the title I'm trying to work out what the chances of drawing $m$ copies of a specific card in $n$ draws are given a deck size of $d$ containing $a$ copies of $A$.

I've tried using permutations for this but must be doing something wrong as the solution i get: \begin{equation} \frac{n!(d-n)!}{m!(n-m)!d!} \end{equation} doesn't even contain $a$, so I must be doing something stupid. Please help a combinatorics noob :(.

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  • $\begingroup$ So you want to draw m cards out of a total and n-m cards out of the remaining d-a cards? $\endgroup$ – barrycarter Feb 5 '16 at 2:51
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Since this is drawing w/o replacement (hypergeometric),

$Pr = \frac{\dbinom{a}{m}\dbinom{n-m}{d-a}}{\dbinom {d}{n}}$

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  • $\begingroup$ Any chance you could talk me through that? $\endgroup$ – Duke of Sam Feb 5 '16 at 7:42
  • $\begingroup$ I have corrected a typo. The deck has $a$ cards of type A and $d-a$ of type B (say). We draw a total of $n$ cards, of which $m$ are of type A, and the remaining $n-m cards from type B. $\endgroup$ – true blue anil Feb 5 '16 at 9:02

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