Calculating the number of poker hands that have exactly 4 different denominations

Question

I'm been working through some examples of questions mentioned in the title and came across this one that I wasn't entirely if I was approaching this correctly.

My initial approach to solving this question was to calculate the probability of drawing such a hand and then multiply this by the total number of hands possible.

To get the probability of such a hand I did this

$$\frac{52}{52} * \frac{48}{51} * \frac{44}{50} * \frac{40}{49} * \frac{12}{48} = 0.05634253701$$

Then to get the total number of possible hands I simply did

$$\binom{52}{5} = 2598960$$

And finally, the possible hands by the probability

$$2598960 * 0.05634253701 = 146432$$

Did I approach this question correctly or is the a simpler way I should be doing a question like this?

Answer 1

Choose the 4 ranks (denominations), which 1 of these ranks appears twice, which 2 suits appear for that rank, and which 1 suit appears for each of the other 3 ranks: $$\binom{13}{4}\binom{4}{1}\binom{4}{2}\binom{4}{1}^3=1,098,240$$

Also known as one pair. An alternative approach is to choose the rank for the pair, the 2 suits for the pair, the other 3 ranks, and the suit for each of the other 3 ranks: $$\binom{13}{1}\binom{4}{2}\binom{12}{3}\binom{4}{1}^3=1,098,240$$

Answer 2

You appear to have tried to evaluate the probability for obtaining four cards of different denominations, and one card from the 12 cards remaining among those denominations. This over counts.

Suppose you select $A\heartsuit,2\clubsuit,3\clubsuit,4\clubsuit$ and then $A\diamondsuit$. But this is the same hand as selecrtng $A\diamondsuit,2\clubsuit,3\clubsuit,4\clubsuit$ and then $A\heartsuit$

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Rather seek the probability for obtaining two from four suits in one from thirteen denominations and one from four suits in each of three from twelve denomination, when selecting any five from all fifty-two cards.