Counting arrangements (permutations)

Goal: Use the multiplication principle and factorials to count ordered arrangements.

Remember: n! means n × (n − 1) × … × 1. P(n, k) = n × (n − 1) × … for k factors.

1. 1.Four different cards (A♠, K♥, Q♦, J♣) are placed in a row. How many different orders are possible?

   Answer: 24

   Working: 4! = 4 × 3 × 2 × 1

2. 2.Three cards are dealt one at a time from a full 52-card deck to three labelled seats. How many different outcomes are there when order and seat matter?

   Answer: 132,600

   Working: 52 × 51 × 50

3. 3.In how many orders can the 13 hearts be arranged in a row?

   Answer: 6,227,020,800

   Working: 13!

4. 4.Calculate P(10, 3).

   Answer: 720

   Working: 10 × 9 × 8

5. 5.The four aces are placed in a row with the ace of spades always first. How many orders are possible?

   Answer: 6

   Working: 3! = 3 × 2 × 1 for the other three aces

6. 6.Two cards are drawn in order from a deck without replacement. How many ordered pairs are possible?

   Answer: 2,652

   Working: 52 × 51

7. 7.20! is about 2.4 × 10^18. Explain why 52! must be far larger than 10^18.

   Answer: 52! continues the product beyond 20! with 32 more factors, each at least 21, so it is vastly bigger. In fact 52! ≈ 8.07 × 10^67.