Permutations (Ordered Selections)

Statement

The permutation formula counts the number of ordered arrangements of objects chosen from a larger set. It is given by:

\[ {}^nP_r = \frac{n!}{(n-r)!} \]

Why it’s true

• There are \(n\) choices for the first object, \((n-1)\) choices for the second, and so on until \(r\) objects are chosen.
• This product is equivalent to \(\frac{n!}{(n-r)!}\), because the remaining \((n-r)\) factors of \(n!\) cancel out.

Recipe (how to use it)

1. Identify the total number of objects (\(n\)) and the number selected (\(r\)).
2. Compute \(n!\).
3. Compute \((n-r)!\).
4. Divide: \(\frac{n!}{(n-r)!}\).

Spotting it

Use permutations when order matters — for example, arranging people in a line, assigning positions, or creating codes.

Common pairings

• Permutations vs combinations (order vs orderless).
• Factorials and probability problems.
• Code/password counting problems.

Mini examples

1. Given: 5 people, choose 2 to arrange in order. Answer: \({}^5P_2 = 5!/3! = 20\).
2. Given: 7 books, choose 3 for a shelf in order. Answer: \({}^7P_3 = 7!/4! = 210\).

Pitfalls

• Using combinations instead of permutations (forgetting that order matters).
• Calculating factorials incorrectly.

Exam strategy

• Underline whether order matters — this decides permutation vs combination.
• Use calculator factorial (!) key where possible.
• Simplify fractions by canceling before multiplying.

Summary

The permutation formula \({}^nP_r = \frac{n!}{(n-r)!}\) counts ordered arrangements of \(r\) objects chosen from \(n\). Use it when order is important.