probability of being dealt four of a kind in poker

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I have to resolve this exercise:

I have 52 cards. I get 5 cards. Calculate the probability I get a poker hand of four-of-a-kind.

well I applied the formula ${52 \choose 5}=\frac{52!}{5!47!}$ to calculate every possible combination.but after here I don't know how can go ahead. Some help?

Edit

I made mistake.Now it's correct

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2 Answers

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Hint: Your formula is the total number of $5$ card hands. How many choices of rank of the four are there? How many choices for the other card?

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According to Wiki:Poker probability, or better this here, you'll have the following:

There are 624 possible hands including four of a kind; the probability of being dealt one in a five-card deal is $\frac {C_{13}^1 C_{4}^4 \cdot C_{12}^1 C_{4}^1} {C_{52}^5} = \frac {13 \cdot 1 \cdot 12 \cdot 4} {2{,}598{,}960} \approx 0.024\% $.

with $C_a^b=\binom{a}{b}$.

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