Example 1
- n = 5
- k = 2
10
What is C(5,2)? 10. Order does not count.
Type n and k. The combination C(n,k) counts ways to choose k items from n when order does not matter. C(5,2) = 10. C(10,3) = 120. These are combinations without repetition.
Combinations without repetition. k cannot be larger than n. C(n,0) = 1 and C(n,n) = 1.
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A combination C(n,k), also the binomial coefficient, is the number of ways to choose k items from n when order does not count. C(5,2) = 10: from five people you pick a pair for a shift, and Anna-Bartek is the same choice as Bartek-Anna.
The formula is n! / (k! · (n-k)!). Factorial n! sits in the numerator, and the two factorials in the denominator strip the orders you do not distinguish. C(5,2) = 120 / (2 · 6) = 10. C(n,k) = C(n, n-k), so C(5,3) is also 10.
The fields are n and k, whole numbers, n ≥ k ≥ 0. The calculator rounds the combination to an integer. When k is larger than n, the result stays blank, because that choice does not exist.
C(n,0) = 1 and C(n,n) = 1: one way to pick nothing and one way to take everyone. That rests on 0! = 1. A variation V(n,k) counts the same items with order, so V(5,2) = 20, not 10.
The header units switch does not convert anything here. 5 and 2 stay C(5,2) = 10 whatever the metric label says. This is a count of sets, not a length.
C(10,3) = 120. C(6,6) = 1. C(8,0) = 1. C(6,2) = 15. C(7,3) = 35. When k > n the calculator does not invent a number. The field stays empty.
C(n,k) = n! / (k! · (n-k)!), for 0 ≤ k ≤ n
A combination C(n,k) = n! / (k!(n-k)!). C(5,2) = 10. C(10,3) = 120. Order does not count.
10
What is C(5,2)? 10. Order does not count.
120
How many sets of 3 from 10? C(10,3) = 120.
1
Is C(8,0) zero? No. One way to choose nothing.
10. The formula 5! / (2! · 3!) = 120 / (2 · 6) = 10. The pair A-B and B-A is one choice.
You pick k distinct items from n and do not put them back. Order does not make a new set. That is this card, C(n,k).
120. C(10,7) is also 120, because C(n,k) = C(n, n-k).
A combination ignores order. A variation keeps it. V(5,2) = 20, C(5,2) = 10, because each set has 2! orders.
Both are 1. One way to pick the empty set and one way to take everyone. 0! = 1 holds that formula.
That choice does not exist. The result is a blank result, not zero and not a fraction.
No. 6! = 720 is the arrangements of all six. C(6,6) = 1, because there is one full set.
n is the size of the collection, k is the size of the choice. From 10 people you take 3, not the other way round.
Yes, to an integer. For a legal n and k the value is already whole. Rounding only guards leftover division noise.
When k = 0 or k = n, one factorial is 0! = 1. Without that convention the formula would break.
The calculator computes the same formula as the definition below.
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