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CombinatoricsHard

From a well-shuffled standard $52$ -card deck, $7$ cards are dealt. What is the probability that all $4$ suits appear among them? Give a fraction in lowest terms.

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Hint

Count the hands that miss some suit by inclusion-exclusion over the missing suits.

Solution

The number of hands is $(7 52) = 133784560$ .
Hands missing a given suit number $(7 39) = 15380937$ , hands missing $2$ given suits number $(7 26) = 657800$ , and hands missing $3$ given suits number $(7 13) = 1716$ .
With $4$ , $6$ and $4$ choices of those suits, the favorable count is $133784560 - 61523748 + 3946800 - 6864 = 76200748$ .
The probability is $76200748 \div 133784560$ , which reduces to $\frac{63713}{111860}$ .

Answer: $\frac{63713}{111860}$

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