1) 5 cards are dealt from a standard 52-card deck. What is the probability that the draws will wield:

a) 3 aces and 2 kings?

b) 3 of a kind and 2 of a kind (a full house)

c) 1 ace, 1 two, 1 three, 1 four, 1 five (a straight)

d) any straight ( 5 ordered cards)

for a) I have that it is $\displaystyle \frac{24}{2598960}$. since it's $\displaystyle \frac{\begin{pmatrix} 4 \\ 2 \end{pmatrix} \begin{pmatrix} 4 \\ 3 \end{pmatrix}}{\begin{pmatrix} 52 \\ 5 \end{pmatrix}}$

for b) all I know is that it's going to be $\displaystyle \frac{3774}{2598960}$

for c) is it $\displaystyle P^{52}_{5} = \frac{52!}{48!}$ for the denominator and $\displaystyle \binom{4}{1}^5$ for the numerator which equals $\displaystyle \frac{1024}{\frac{52!}{48!}}$

and d) I think that the denominator is the same as c) but the numerator will be 9 times the original, since there are 9 different choices which equals $\displaystyle \frac{(9)(1024)}{\frac{52!}{48!}}$

are these right?