Combinatorics: Ways to pick 5 card poker hands to get the ace of spaces and another the ace of hearts.

Question

Please someone explain to me how to do the following problem:

How many different 5 card poker hands can you get where one of the cards must be the ace of spades and another the ace of hearts?

My answer is $$\binom{52}{4} + \binom{51}{3}$$ because there are total of $52$ cards with $4$ aces. There are $\binom{52}{4}$ ways to choose the ace of spade. After we choose the ace of spade, there are total of $51$ cards left in the deck with $3$ aces. There are $\binom{51}{3}$ ways to choose the ace of hearts. Then we add them both.

Is my answer right?

Answer 1

Once you have selected the ace of spades and the ace of hearts, you have 50 remaining cards from which you choose 3 to complete the 5-card hand. Thus the total number of ways is simply $$\binom{50}{3}.$$

Answer 2

There is only one ace of spades and one ace of hearts
Take the other three from the 50 remaining

$$\frac{\binom{2}{2} \binom{50}{3}} {\binom{52}{5}} = 1 / 132.6 \approx 0.0075414781 $$