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I have a question.

I'm looking to calculate the probability of getting $4$ different suits cards in a $5$ card poker game using a standard $52$ card deck.

I think this is: $$\frac{\dbinom{4}{1}\dbinom{13}{2}\dbinom{13}{1}^{3}}{\dbinom{52}{5}}$$

What do you think? Thanks.

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  • $\begingroup$ Relevant: math.stackexchange.com/questions/211653/… $\endgroup$ – Vepir May 8 '16 at 7:54
  • $\begingroup$ Hey,I deal $5$cards. $\endgroup$ – NM2 May 8 '16 at 7:56
  • $\begingroup$ Do you have and idea why this answer isnt correct ? $\frac{52}{52}\cdot \frac{39}{51} \cdot \frac{26}{50} \cdot \frac{13}{49} \cdot \frac{48}{48}$ $\endgroup$ – NM2 May 8 '16 at 8:01
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The answer you have quoted in the body of the question is correct.

The answer you have posted in the comment is incorrect, because:

The first $4$ cards are permutations of $ABCD$ while the last is simply appended.

To make the numerator a true permutation, the last card could be placed at any of five spots,$\;(\;\bullet A \bullet B \bullet C \bullet D \bullet\;)$ and you would need to divide by $2$ to correct for two identical suits

ADDED

The link Matta has given is for four cards dealt, where the two methods would tally effortlessly:

$$\frac{13^4}{\binom{52}4} = \frac{52}{52}\cdot\frac{39}{51}\cdot\frac{26}{50}\cdot\frac{13}{49}$$

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