1 · IB Maths AA
Permutations and combinations
Pick n objects and r of them, then step through every ordered arrangement. Each row of the grid is one unordered selection with its r! orderings, which shows why nCr = nPr ÷ r!.
nPr = n! / (n − r)!
nCr = n! / (r! (n − r)!)
nCr = nPr / r!
nCr = n! / (r! (n − r)!)
nCr = nPr / r!
Try this
- Set n = 4 and r = 2. Count the rows and the cells in each row, then check that 12 = 6 × 2.
- With n = 7, compare r = 2 with r = 5. Why is nCr the same for both while nPr is very different?
- Set r = n: there is one selection but n! arrangements. Then set r = 0 and explain why both counts equal 1.
Exam tip: decide first whether order matters (positions, codes and rankings use nPr; teams, committees and hands use nCr), and when a question needs both, multiply the number of selections by the number of ways to order each one.