Questions: This exercise refers to a standard deck of playing cards. Assume that 8 cards are randomly chosen from the deck. How many hands contain exactly two 8s and two 9s? hands

Solution Steps

To solve this problem, we need to calculate the number of ways to choose exactly two 8s and two 9s from a standard deck of 52 cards. A standard deck has 4 cards of each rank. We will use combinations to determine the number of ways to choose 2 cards from the 4 available 8s and 2 cards from the 4 available 9s. The total number of such hands is the product of these two combinations.

To find the number of ways to choose 2 cards from the 4 available 8s, we use the combination formula:

\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \]

Similarly, to find the number of ways to choose 2 cards from the 4 available 9s, we again use the combination formula:

\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \]

The total number of hands containing exactly two 8s and two 9s is the product of the two combinations calculated:

\[ \text{Total Hands} = \binom{4}{2} \times \binom{4}{2} = 6 \times 6 = 36 \]

Final Answer