If a 5-card hand is dealt from a standard 52 card deck, how many hands will contain exactly 2 aces? (not probability!)
The correct answer and explanation is:
To calculate the number of 5-card hands that contain exactly 2 aces, the problem can be broken down into two parts: choosing the 2 aces and selecting the remaining 3 cards from the non-ace cards.
Step 1: Choosing the 2 aces
A standard deck has 4 aces (one for each suit: hearts, diamonds, clubs, and spades). The number of ways to select 2 aces from these 4 is calculated using the combination formula: Combinations of 2 aces=(42)=4×32×1=6\text{Combinations of 2 aces} = \binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6
So, there are 6 different ways to choose 2 aces from the 4 aces.
Step 2: Choosing the remaining 3 cards
After choosing 2 aces, there are 50 cards left in the deck (since 4 aces were originally in the deck). Out of these 50 cards, 3 must be selected for the hand. These remaining cards cannot be aces, so we are choosing from the 48 non-ace cards. The number of ways to select 3 cards from these 48 non-ace cards is given by: Combinations of 3 non-ace cards=(483)=48×47×463×2×1=17296\text{Combinations of 3 non-ace cards} = \binom{48}{3} = \frac{48 \times 47 \times 46}{3 \times 2 \times 1} = 17296
So, there are 17,296 ways to select the remaining 3 non-ace cards.
Step 3: Total number of hands
To find the total number of hands containing exactly 2 aces, multiply the number of ways to choose the aces by the number of ways to choose the remaining cards: Total hands with 2 aces=(42)×(483)=6×17296=103776\text{Total hands with 2 aces} = \binom{4}{2} \times \binom{48}{3} = 6 \times 17296 = 103776
Therefore, there are 103,776 possible 5-card hands that contain exactly 2 aces.