Problem 16 - Algebra for College Students

Question

How many two-card hands can be dealt from a deck of 52 playing cards? 1326

Step-by-Step Solution

Answer

There are 1326 different two-card hands possible from a deck of 52 cards.

Step 1: Understanding the Problem

We want to find out how many different two-card combinations can be made from a standard deck of 52 playing cards. A combination does not consider the order of the cards, so "Ace of Spades and King of Hearts" is the same as "King of Hearts and Ace of Spades."

Step 2: Using the Combination Formula

To find the number of combinations, we use the combination formula \( C(n, k) = \frac{n!}{k!(n-k)!} \), where \( n \) is the total number of items to choose from, \( k \) is the number of items to choose, and \( ! \) denotes factorial.

Step 3: Substitute Values into the Formula

For this problem, \( n = 52 \) and \( k = 2 \). Substituting these values into the combination formula, we have \( C(52, 2) = \frac{52!}{2!(52-2)!} = \frac{52!}{2! \, 50!} \).

Step 4: Simplify the Expression

Simplify \( \frac{52!}{2! \, 50!} \) by canceling \( 50! \) from the numerator and denominator, yielding \( \frac{52 \times 51}{2!} = \frac{52 \times 51}{2} \).

Step 5: Calculate the Result

Perform the multiplication and division: \( 52 \times 51 = 2652 \) and \( \frac{2652}{2} = 1326 \).

Step 6: Conclusion

Therefore, there are 1326 different possible two-card hands that can be dealt from a standard deck of 52 cards.

Key Concepts

Combination Formula

To figure out how many different ways you can choose a set of items from a larger group, we use the Combination Formula. Unlike permutations, combinations focus on choosing items without considering the order. In this formula, given a total number of items 'n,' and you want to choose 'k' items, the combination is expressed as \( C(n, k) = \frac{n!}{k!(n-k)!} \). This formula gives the number of combinations possible:

• \( n! \) is the factorial of 'n', which signifies the product of all positive integers up to 'n'.
• \( k! \) is the factorial of 'k', representing the number of arrangements of the items we wish to choose.
• \( (n-k)! \) accounts for the remaining items that are not chosen.

By plugging values into this formula, we can calculate how many different sets of 'k' items can be chosen from 'n' total items. It's important to realize the significance that the order does not matter, making combinations ideal for determining sets like our two-card hand scenario.

Factorial

Factorials are fundamental in combinatorics and are designated by the exclamation point (!). If you're unfamiliar, here’s a simple breakdown: a factorial of a non-negative integer 'n', denoted as \( n! \), means the product of all positive integers less than or equal to 'n'. For instance, \( 5! \) equals \( 5 \times 4 \times 3 \times 2 \times 1 = 120 \). Factorials grow rapidly with larger numbers, which makes them useful for counting large sets of combinations or permutations without listing every possibility. They simplify the math when calculating the number of ways to arrange a set of items.

Card Combinations

Card combinations are a classic example of using combinatorial mathematics in real-life scenarios. When dealing two cards from a complete deck of 52 cards, the Combination Formula becomes pivotal. In this problem, where order doesn't matter, we calculate the number of ways to select 2 cards from 52:1. Identify 'n' as 52 and 'k' as 2, representing the total cards and selected cards.2. Apply the formula \( C(52, 2) = \frac{52!}{2!(52-2)!} = \frac{52!}{2! \, 50!} \).3. Continue by simplifying \( \frac{52 \times 51}{2!} = \frac{52 \times 51}{2} \).Each part of the formula works together to exclude the order and focus on quantity, proving that there are 1326 different two-card hands possible. Working through this example provides insight into the power and necessity of combinatorics in evaluating and predicting outcomes in otherwise complex counting problems.