Permutations of r Objects from n
Simple Explanation
More generally, the number of ways to choose AND arrange (in order) just r objects out of a total of n distinct objects is P(n,r) = n! / (n−r)!.
Why Do We Need It?
Many real counting problems only involve arranging PART of a larger set (like awarding the top 3 medals out of many competitors) — this formula handles that general case directly.
Formula
Permutations of r Objects from n
P(n,r) = n! / (n−r)!
The number of ways to choose AND arrange r objects, in order, out of a total of n distinct objects.
- n
- — the total number of distinct objects available
- r
- — the number of objects being chosen and arranged, r ≤ n
When to use it: Whenever the ORDER of the chosen objects matters (e.g. 1st, 2nd, 3rd place).
Worked Example
Count ordered selections from a larger set
In how many ways can 3 medals (gold, silver, bronze) be awarded to 8 runners?
Why Does This Work?
There are 8 choices for who gets gold, then 7 remaining choices for silver, then 6 remaining for bronze — 8×7×6, by the Fundamental Counting Principle. Dividing 8! by 5! (which is 8×7×6×5×4×3×2×1 divided by 5×4×3×2×1) leaves exactly this same product, 8×7×6.
Real-Life Example
Assigning distinct roles from a pool of candidates
A company needs to fill 3 different specific roles (manager, assistant, and treasurer) from a pool of 10 qualified candidates.
Since each role is distinct (order matters — being manager is different from being treasurer), this is exactly a permutations calculation, P(10,3).
Practice
In how many ways can a president and vice-president be chosen from 10 candidates?
MediumCommon mistake
Using combinations instead when order actually matters (like distinct medal positions or job titles) — permutations count arrangements, combinations do not.
Quick Review
- P(n,r) = n! / (n−r)!.
- Use whenever ORDER matters among the r chosen objects.
- A special case of this formula (r=n) reduces to the simpler P(n,n)=n! from before.