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What is the probability of getting the following hand? (Red 10, Black J, Red Q, Black K and Red A) or (Black 10, Red J, Black Q, Red K, Black A), we do not consider ordering in this case.

This is my solution $$(10/52 \cdot 8/51 \cdot 6/50 \cdot 4/49 \cdot 2/48) \cdot 2$$ Just wondering am I on the right track. Thanks a lot!

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    $\begingroup$ It is correct, though there are other approaches which give the same answer. Personally I would have written $2 \times \dfrac{5! \times 2^5}{52\times 51 \times 50 \times 49 \times 48}$ $\endgroup$ – Henry Mar 21 at 8:45
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Yes, it is correct. There are $2\cdot 2\cdot 2\cdot 2\cdot 2=2^5$ hands of the first type. The same number for the second type. There is no hand which is both of the first and of the second type. Then the required probability is $$\frac{2^5+2^5}{\binom{52}{5}}$$ which is the same of yours.

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