10C3 committees
Input
- n: 10
- r: 3
Calculation
10!/(3!7!) = 120
Result
120 combinations.
Calculate nPr and nCr for selection counting problems in probability and combinatorics.
Overview
Combinatorics counts arrangements. Permutations nPr count ordered selections; combinations nCr count unordered selections. Password ordering and committee selection illustrate the distinction.
Formulas use factorials: nPr = n!/(n−r)! and nCr = n!/[r!(n−r)!]. Factorials grow rapidly, so calculators use cancellation to avoid overflow.
Order matters for permutations: ABC differs from BAC. Order does not matter for combinations: {A,B,C} is one committee regardless of listing order.
Identify whether items are distinct, whether repetition is allowed, and whether order matters before choosing nPr versus nCr. Misclassification is the top student error in counting problems.
Steps
Total items available.
Items chosen.
Ordered versus unordered.
Apply factorial formula with cancellation.
Cannot choose more items than exist.
Confirm order and repetition assumptions.
Math
nPr = n!/(n−r)!; nCr = n!/[r!(n−r)!]
Permutations count arrangements; combinations divide out r! redundant orderings.
Worked cases
10!/(3!7!) = 120
120 combinations.
10!/7! = 720
720 permutations.
Large combination count
2,598,960 distinct hands.
Why use this
Questions