Question

What is the answer to the following question and how do you solve it? "This exercise refers to a standard deck of playing cards. Assume that 7 cards are randomly chosen from the deck. How many hands contain exactly two 2s and two 9s?"

Answer 1

To find the number of hands with exactly two 2s and two 9s when choosing 7 cards from a standard deck, multiply the combinations of selecting 2s and 9s by the combinations of selecting the remaining 3 cards, yielding 441,072 possible hands.

Step-by-step explanation:

To determine the number of hands that contain exactly two 2s and two 9s from a standard deck of 52 playing cards, we use combinatorics. There are four 2s in the deck and four 9s in the deck. So, the number of ways to choose two 2s from the four available is combin(4,2), which is 6. Similarly, the number of ways to choose two 9s from the four available is also combin(4,2), which is 6. Now we have 7-4=3 cards left to choose, and there are 44 cards remaining in the deck (since 4 of the cards are already two 2s and two 9s). Therefore, the number of ways to choose the remaining three cards is combin(44,3), which is 12,276.

The total number of hands containing exactly two 2s and two 9s is the product of the combinations:

6 (for the 2s) × 6 (for the 9s) × 12,276 (for the remaining three cards) = 441,072.