Question 1192165: How many ways are there to choose two twos from a standard 52-card deck?
How many ways are there to choose three cards from a standard 52-card deck without choosing any twos?
How many five-card hands (drawn from a standard 52-card deck) contain exactly two twos?
How many five-card hands (drawn from a standard 52-card deck) contain a two-of-a-kind?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's break down each part of the problem:
**1. Choosing Two Twos:**
* There are four twos in a standard 52-card deck.
* We want to choose two of them.
* The number of ways to do this is given by the combination formula "4 choose 2":
⁴C₂ = 4! / (2! * 2!) = (4 * 3) / (2 * 1) = 6
**2. Choosing Three Cards Without Any Twos:**
* There are 48 cards in the deck that are *not* twos (52 total cards - 4 twos).
* We want to choose three of these 48 cards.
* The number of ways to do this is:
⁴⁸C₃ = 48! / (3! * 45!) = (48 * 47 * 46) / (3 * 2 * 1) = 17296
**3. Five-Card Hands with Exactly Two Twos:**
* First, we choose two twos (as calculated in part 1): ⁶C₂ = 6 ways.
* Then, we need to choose the remaining three cards from the 48 non-two cards (as calculated in part 2): ⁴⁸C₃ = 17296 ways.
* To get the total number of five-card hands with exactly two twos, we multiply these two results together:
6 * 17296 = 103776
**4. Five-Card Hands with a Two-of-a-Kind:**
This is trickier because it includes more than just two twos. It means *any* pair (two of the same rank) but not three or four of a kind.
* **Choose the rank for the pair:** There are 13 ranks in a deck (Ace through King). We choose one of these ranks: ¹³C₁ = 13 ways.
* **Choose the two cards for the pair:** There are four cards of the chosen rank. We choose two of them: ⁴C₂ = 6 ways.
* **Choose the ranks for the other three cards:** We need three more cards, each of a different rank, and these ranks must be different from the rank we chose for the pair. There are 12 remaining ranks, and we choose three of them: ¹²C₃ = 220 ways.
* **Choose the suit for each of the three remaining cards:** For each of the three cards, we have 4 suit options. This gives us 4 * 4 * 4 = 4³ = 64 possibilities.
* **Multiply everything together:** 13 * 6 * 220 * 64 = 1098240
Therefore, there are 1,098,240 five-card hands with exactly one pair (two of a kind).
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