Permutations and Combinations Count
Permutations and Combinations Count
Counting how many ways a set of items can be arranged or chosen is a foundational skill behind probability calculations, scheduling problems, lottery odds, and reliability analysis — anywhere the question "how many possible outcomes are there?" comes up. Permutations count arrangements where order matters; combinations count selections where it does not.The two numbers can differ by orders of magnitude for the same n and r, and mixing them up is a classic source of error — for example, computing the odds of winning a lottery requires combinations (the order the balls are drawn does not matter for winning), while ranking the top three finishers in a race requires permutations.
The number of permutations is nPr = n!/(n−r)! and the number of combinations is nCr = n!/(r!(n−r)!). where n_tot is the total number of items and r_pick is the number chosen or arranged.
Inputs
n_tot := 10=10
r_pick := 3=3
Dividing n! by (n−r)! cancels all the factors below n−r, leaving exactly the count of ordered arrangements of r items.
nPr_val := factorial(n_tot)/factorial(n_tot-r_pick)=720
Dividing the permutation count further by r! removes the ordering within each selected group, leaving the count of unordered selections.
nCr_val := factorial(n_tot)/(factorial(r_pick)·factorial(n_tot-r_pick))=120
Results
With 10 items and 3 chosen, there are 720 ways to arrange them in order but only 120 ways to select them without regard to order — exactly 3! = 6 times fewer, since each combination corresponds to 6 orderings. Whenever "how many ways" is the question, the first step is always deciding whether order matters; getting that wrong gives an answer off by a factor of r!.
Assumptions & limitations
- Items are distinguishable.
- Selection without replacement.
- r_pick must not exceed n_tot.
- No items are identical (no repeated-element correction needed).