How this instrument works
C(n, r), read n choose r, counts how many distinct r-item subsets sit inside a set of n distinct items when the order of selection carries no meaning. The factorial formula n! ⁄ (r!(n−r)!) comes from a short argument: line up all n items in some order, which can be done n! ways, then stop caring about the internal order of the first r (divide by r!) and of the remaining n−r (divide by (n−r)!). What is left counts only which items landed in the first group, and that is exactly a subset.
Every entry equals the sum of the two entries above it in Pascal's triangle: C(n,r) = C(n−1,r−1) + C(n−1,r), since a chosen item either includes one particular member or excludes it, and the two cases cover every possibility. The triangle carries Pascal's name in the West, yet it appears centuries earlier — in the Persian mathematician Al-Karaji's work around the year 1000, and in China with Jia Xian in the eleventh century, later set out fully by Yang Hui. A second identity follows just as directly: C(n,r) = C(n,n−r), because choosing r items to include is the same act as choosing n−r items to leave out.
Two edge cases anchor the whole table. Choosing zero items from any set has exactly one outcome, the empty selection, so C(n,0) = 1 for every n; choosing all n items is likewise the single outcome of taking everything, so C(n,n) = 1. A more surprising fact sits one step further out: add every entry across row n and the total is 2^n, the count of every subset of any size at once — the same 2^n that counts every possible yes-or-no vote across n people, tying one row of Pascal's triangle directly to binary choice.
- Enter the size of the full set in n — the total number of distinct items available to choose from.
- Enter the size of the group being selected in r — how many of those items you are picking at once.
- Read n choose r for the count of distinct subsets, with the order of selection ignored entirely.
- Swap r for n minus r and run it again; n choose r returns the identical figure, confirming the symmetry.
- Entering an r larger than n triggers a flag — no set of size n contains a subset bigger than itself.
Worked example — a five-person shortlist, pick two
A hiring panel has narrowed a search to 5 candidates and needs to send exactly 2 of them forward for final interviews, with no distinction between the two forwarded slots. Set n = 5 and r = 2: n choose r = 5! ⁄ (2!·3!) = 120 ⁄ (2×6) = 120 ⁄ 12 = 10. There are 10 different pairs the panel could forward, and every one of them is an equally valid choice of 2 candidates from the 5.
The symmetry rule checks the answer for free: choosing 2 candidates to forward is the same act as choosing the other 3 to set aside, so n choose r must equal 5 choose 3. Working that side out gives 5! ⁄ (3!·2!) = 120 ⁄ (6×2) = 10 as well — both routes land on the same 10, which is exactly what C(n,r) = C(n,n−r) predicts.
Questions
What does n choose r actually count?
It counts the number of distinct ways to select r items from a set of n distinct items when the order of selection does not matter. Picking Alex-then-Sam for a two-person team counts the same as picking Sam-then-Alex — both produce the identical pair, so the formula divides out every internal ordering with r! in the denominator.
How does a combination differ from a permutation?
A permutation counts ordered arrangements; a combination, which is what n choose r gives, ignores order entirely. The two are linked by P(n,r) = r! × C(n,r): take every combination and multiply by the r! ways to order its members, and you recover the permutation count. Choosing a 2-person team from 5 people gives 10 combinations, but 20 ordered pairs.
Why does C(n,r) equal C(n, n−r)?
Because choosing which r items to include is the same decision as choosing which n−r items to leave out — naming one side of the split automatically names the other. Five choose two equals five choose three for exactly this reason: naming 2 committee members also names the 3 people left off the committee, so both counts must agree.
What happens when r is 0 or equal to n?
Both give exactly 1. Choosing zero items has one outcome, the empty selection, so C(n,0) = 1 for every n. Choosing every item also has one outcome, taking the whole set, so C(n,n) = 1. These two values sit at the ends of every row of Pascal's triangle, and this sheet returns them the same way it handles any other pair.
Where does the name binomial coefficient come from?
From the binomial theorem, which expands (x + y)^n into a sum of terms, and the coefficient in front of x^(n−r) y^r turns out to be exactly C(n,r). Multiply out (x + y)^5 and the coefficients that appear — 1, 5, 10, 10, 5, 1 — are precisely row 5 of Pascal's triangle, the same numbers this calculator returns for n = 5.
What if I enter an r larger than n?
There are no r-item subsets of a set smaller than r, so this sheet flags the input rather than showing a result. Mathematically the coefficient is treated as 0 in that region — you cannot choose 6 items from a group of 5 — but the practical rule for this calculator is simply that r must stay at or below n.