<s>[INST] What is the probability of being dealt a hand where all $5$ cards are different denominations? In a game of poker you are dealt $5$ cards at random from a standard deck of $52$. A standard deck of $52$ contains $13$ different denominations of cards (Ace, 2, 3, ..., 10, jack, queen, king), each in four different suits
(a) What is the probability of being dealt a hand where all $5$ cards are different denominations?
(b) What is the probability of being dealt a hand that includes at least one Ace?
My attempts:
(a) Let D = the event where a hand is dealt such that all $5$ cards are different denominations.
There are ($\binom{13}{5}$ ways to pick the denominations $\times \binom{4}{1}$ ways to pick the suit)/ ($\binom{52}{5}$ total combinations)
(b) Let A = the event where a hand is dealt such that it includes at least one ace.
$P(a) =$ ($\binom{4}{1}$ ways to pick the ace $\times \binom{13}{4}$ ways to pick the other cards $\times \binom{4}{1}$ ways to pick the suit)/ ($\binom{52}{5}$ total combinations)
If someone could walk me through a problem similar to this it'd be really appreciated this stuff is sort of confusing. (poor cs major)
 [/INST] 
What is the probability of being dealt a hand where all five cards are of different denominations?

There are $\binom{52}{5}$ ways to select a five-card hand.
There are $\binom{13}{5}$ ways to select five different denominations.  For each such denomination, there are $\binom{4}{1}$ ways to choose the suit of the card of that denomination.  Therefore, there are $\binom{13}{5}\binom{4}{1}^5$ favorable cases.  
Hence, 
$$\Pr(\text{five different denominations}) = \frac{\dbinom{13}{5}\dbinom{4}{1}^5}{\dbinom{52}{5}}$$
As Lord Shark the Unknown indicated in the comments, you failed to pick a suit for each denomination.

What is the probability of being dealt a hand with at least one ace?

Method 1:  Since there are four aces in the deck, such a hand either contains one ace and four other cards, two aces and three other cards, three aces and two other cards, or four aces and one other card.  Since there are $4$ aces and $48$ other cards in the deck,
$$\Pr(\text{at least one ace}) = \frac{\dbinom{4}{1}\dbinom{48}{4} + \dbinom{4}{2}\dbinom{48}{3} + \dbinom{4}{3}\dbinom{48}{2} + \dbinom{4}{4}\dbinom{48}{1}}{\dbinom{52}{5}}$$
Method 2: We subtract the probability of selecting no aces from $1$.  If no aces are selected, we must select five of the other $48$ cards.  Hence,
$$\Pr(\text{no aces}) = \frac{\dbinom{48}{5}}{\dbinom{52}{5}}$$
Thus, 
$$\Pr(\text{at least one ace}) = 1 - \frac{\dbinom{48}{5}}{\dbinom{52}{5}}$$
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