Math Problem Statement

In a card game you are dealt seven cards (click image to enlarge). Determine how many different ways you can fan out your hand.:

Now suppose three of the cards in your hand are Hearts; two of the cards are Clubs; and two of the cards are Spades. Keeping the cards together by their suit, determine how many different ways you can fan out your hand.:

Solution

Let's solve this problem step by step.

Part 1: Total Number of Ways to Fan Out the Cards (No Suit Restrictions)

When you are dealt 7 cards, you can arrange them in different ways. The number of different ways to fan out the cards is equivalent to the number of ways to permute 7 distinct objects.

The number of possible arrangements of nn distinct objects is given by the formula for permutations:

P(n)=n!P(n) = n!

For n=7n = 7 cards:

P(7)=7!=7×6×5×4×3×2×1=5040P(7) = 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040

So, the number of different ways to fan out your 7 cards is:

5040 ways5040 \text{ ways}


Part 2: Grouping Cards by Suit (Hearts, Clubs, Spades)

Now, you are given that:

  • 3 of the cards are Hearts,
  • 2 of the cards are Clubs,
  • 2 of the cards are Spades.

You are asked to determine how many different ways you can fan out your hand, keeping the cards together by their suit.

  1. Step 1: Permute the suits themselves: Since we have 3 groups of suits (Hearts, Clubs, Spades), we can permute these groups. The number of ways to arrange 3 groups is:

3!=3×2×1=63! = 3 \times 2 \times 1 = 6

  1. Step 2: Permute the cards within each suit:
    • For the Hearts, we have 3 cards, so they can be arranged in:

3!=3×2×1=6 ways3! = 3 \times 2 \times 1 = 6 \text{ ways}

  • For the Clubs, we have 2 cards, so they can be arranged in:

2!=2×1=2 ways2! = 2 \times 1 = 2 \text{ ways}

  • For the Spades, we have 2 cards, so they can be arranged in:

2!=2×1=2 ways2! = 2 \times 1 = 2 \text{ ways}

  1. Step 3: Multiply the results: To get the total number of ways to fan out the cards while keeping the suits together, we multiply the number of ways to permute the suits by the number of ways to arrange the cards within each suit:

3!×3!×2!×2!=6×6×2×2=144 ways3! \times 3! \times 2! \times 2! = 6 \times 6 \times 2 \times 2 = 144 \text{ ways}


Final Answer

  • Without restrictions (part 1): There are 5040 ways to fan out your 7 cards.
  • Keeping cards together by suit (part 2): There are 144 ways to fan out your 7 cards.

Would you like a detailed explanation of any part? Let me know!


Here are 5 related questions you could explore:

  1. How many ways can you arrange 4 cards, each of a different suit?
  2. What if you had 4 cards of one suit and 3 cards of another? How would that affect the number of ways to fan out your hand?
  3. How many ways can you select 3 cards from a standard deck of 52 cards, all of which are Hearts?
  4. What if you had a hand of 7 cards with all different suits? How would you calculate the permutations then?
  5. How would the calculation change if you were dealt 8 cards, with 4 Hearts, 2 Clubs, and 2 Spades?

Tip: When dealing with permutations of grouped objects, first count the ways to arrange the groups themselves, then calculate the ways to arrange items within each group!

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Math Problem Analysis

Mathematical Concepts

Permutations
Factorial
Combinatorics

Formulas

P(n) = n! (Permutations of n distinct objects)
3! (Permutations of suits)
3! (Permutations within Hearts)
2! (Permutations within Clubs)
2! (Permutations within Spades)

Theorems

Fundamental Principle of Counting

Suitable Grade Level

Grades 10-12