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Counting arrangements: permutations

Why 52 cards can be ordered in more ways than there are atoms on Earth, and how to count ordered selections.

The multiplication principle

If one choice can be made in a ways and a second, independent choice in b ways, then the pair can be made in a × b ways. This simple rule is the engine behind almost every counting formula.

Suppose you lay out three different cards in a row. There are 3 choices for the first position, then 2 remaining cards for the second, then 1 for the last: 3 × 2 × 1 = 6 arrangements.

Factorials

Multiplying all whole numbers from n down to 1 is called n factorial and is written n!. It counts the ways to put n distinct objects in order. By convention 0! = 1, because there is exactly one way to arrange nothing.

A full deck therefore has 52! orderings, which is roughly 8.07 × 10^67. A well-shuffled deck is almost certainly in an order that has never occurred before in the history of card shuffling.

n! = n × (n − 1) × … × 2 × 1 5! = 120 52! ≈ 8.07 × 10^67

Ordered selections of some of the items

Often you do not arrange everything, only k of n items, and the order matters. The first position has n choices, the second n − 1, and so on for k positions. This count is called a permutation of k items from n.

For example, the number of ways to deal five cards to five labelled players, one card each, is 52 × 51 × 50 × 49 × 48 = 311,875,200.

P(n, k) = n! / (n − k)! = n × (n − 1) × … × (n − k + 1) P(52, 5) = 311,875,200

Common mistakes

  • Using permutations when order does not matter. If the five cards are just a set, use combinations instead (next lesson).
  • Forgetting that choices shrink without replacement: the second card has 51 possibilities, not 52.
  • Expecting a calculator to show exact huge numbers. Ordinary floating-point numbers lose precision beyond about 15 digits; this site uses exact big integers.

Check your understanding

How many different orders can the four aces be placed in a row?

4! = 24.

How many ordered pairs (first card, second card) can be drawn from a deck?

P(52, 2) = 52 × 51 = 2,652.

Why is 0! defined as 1?

There is exactly one way to arrange zero objects (do nothing), and the definition keeps formulas such as C(n, 0) = 1 working.