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Math A

Counting: Permutations and Combinations Practice

The key question in counting is whether order matters and whether things are distinguishable. Be clear about what you are counting before applying a formula.

Basic, Standard: Grade 10 · Term 1 / Advanced: Grade 12 · Exam prep

Math problem generator

Level

Permutations and combinations

Formulas
nPr=n(n−1)⋯(n−r+1)=n!(n−r)!,nCr=nPrr!=n!r!(n−r)!{}_{n}\mathrm{P}_{r} = n(n - 1)\cdots(n - r + 1) = \frac{n!}{(n - r)!},\qquad {}_{n}\mathrm{C}_{r} = \frac{{}_{n}\mathrm{P}_{r}}{r!} = \frac{n!}{r!(n - r)!}

Use permutations P\mathrm{P} when order or roles matter (president and vice-president) and combinations C\mathrm{C} when only the selection matters (choosing a committee). With repetition allowed, there are nrn^r arrangements.

Circular arrangements, repeated items, shortest paths

Special arrangements

Circular: (n−1)!(n - 1)! Necklace: (n−1)!2\dfrac{(n - 1)!}{2} With repeated items: n!p! q! r!\dfrac{n!}{p!\,q!\,r!}

Items that must be together are treated as one block; items that must be apart are placed into the gaps after arranging the others. Shortest grid paths are arrangements of right and up moves: mm across and nn up give m+nCm {}_{m+n}\mathrm{C}_{m} paths.

Grouping and combinations with repetition

Putting 6 people two per room into rooms A, B and C gives 6C2×4C2=90 {}_{6}\mathrm{C}_{2} \times {}_{4}\mathrm{C}_{2} = 90 ways, but splitting them into three unlabeled pairs gives 90÷3!=1590 \div 3! = 15. Divide by the arrangements of groups of equal size when the groups are unlabeled.

Combinations with repetition

Nonnegative integer solutions of x+y+z=nx + y + z = n correspond to arrangements of nn circles and 2 bars, so there are n+2C2 {}_{n+2}\mathrm{C}_{2} of them.