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Combination N Choose K

Tool to generate combinations. In mathematics, a choice of k elements out of n distinguishable objects (k choose n), where the order does not matter, is represented by a list of elements, which cardinal is the binomial coefficient.

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Combination N Choose K -

Tag(s) : Combinatorics

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Combination N Choose K

Combinations Generator






Number of Combinations Calculator



See also: Permutations

Combinations and Lottery Games

To get a list of combinations with a guaranteed minimum of numbers (also called reduced lottery draw), dCode has a tool for that:

To draw random numbers (Lotto, Euromillions, Superlotto, etc.)

Tool to generate combinations. In mathematics, a choice of k elements out of n distinguishable objects (k choose n), where the order does not matter, is represented by a list of elements, which cardinal is the binomial coefficient.

Answers to Questions

How to generate combinations of n choose k?

The generator allows to select values of $ k $ and $ n $, and generates list of combinations with digits or letters (or a custom list).

Example: 4 choose 2 generates: (1,2),(1,3),(1,4),(2,3),(2,4),(3,4)

The generation is limited to 2000 lines. Combinatorics can introduce huge numbers, this limits secures the computation server.

To generates large lists, dCode can generate them upon (paid) request.

How to count the number of combinations of n choose k?

The calculation uses the binomial coefficient: $$ C_n^k = \binom{n}{k} = \frac{n!}{k!(n-k)!} $$

Combinations uses calculus of factorials (the exclamation mark: !).

3 choose 2 = 3 combinations(1,2)(1,3)(2,3)
4 choose 2 = 6 combinations(1,2)(1,3)(1,4)(2,3)(2,4)(3,4)
5 choose 2 = 10 combinations(1,2)(1,3)(1,4)(1,5)(2,3)(2,4)(2,5)(3,4)(3,5)(4,5)
6 choose 2 = 15 combinations(1,2)(1,3)(1,4)(1,5)(1,6)(2,3)(2,4)(2,5)(2,6)(3,4)(3,5)(3,6)(4,5)(4,6)(5,6)
7 choose 2 = 21 combinations(1,2)(1,3)(1,4)(1,5)(1,6)(1,7)(2,3)(2,4)(2,5)(2,6)(2,7)(3,4)(3,5)(3,6)(3,7)(4,5)(4,6)(4,7)(5,6)(5,7)(6,7)
8 choose 2 = 28 combinations(1,2)(1,3)(1,4)(1,5)(1,6)(1,7)(1,8)(2,3)(2,4)(2,5)(2,6)(2,7)(2,8)(3,4)(3,5)(3,6)(3,7)(3,8)(4,5)(4,6)(4,7)(4,8)(5,6)(5,7)(5,8)(6,7)(6,8)(7,8)
9 choose 2 = 36 combinations(1,2)(1,3)(1,4)(1,5)(1,6)(1,7)(1,8)(1,9)(2,3)(2,4)(2,5)(2,6)(2,7)(2,8)(2,9)(3,4)(3,5)(3,6)(3,7)(3,8)(3,9)(4,5)(4,6)(4,7)(4,8)(4,9)(5,6)(5,7)(5,8)(5,9)(6,7)(6,8)(6,9)(7,8)(7,9)(8,9)
4 choose 3 = 4 combinations(1,2,3)(1,2,4)(1,3,4)(2,3,4)
5 choose 3 = 10 combinations(1,2,3)(1,2,4)(1,2,5)(1,3,4)(1,3,5)(1,4,5)(2,3,4)(2,3,5)(2,4,5)(3,4,5)
6 choose 3 = 20 combinations(1,2,3)(1,2,4)(1,2,5)(1,2,6)(1,3,4)(1,3,5)(1,3,6)(1,4,5)(1,4,6)(1,5,6)(2,3,4)(2,3,5)(2,3,6)(2,4,5)(2,4,6)(2,5,6)(3,4,5)(3,4,6)(3,5,6)(4,5,6)
7 choose 3 = 35 combinations(1,2,3)(1,2,4)(1,2,5)(1,2,6)(1,2,7)(1,3,4)(1,3,5)(1,3,6)(1,3,7)(1,4,5)(1,4,6)(1,4,7)(1,5,6)(1,5,7)(1,6,7)(2,3,4)(2,3,5)(2,3,6)(2,3,7)(2,4,5)(2,4,6)(2,4,7)(2,5,6)(2,5,7)(2,6,7)(3,4,5)(3,4,6)(3,4,7)(3,5,6)(3,5,7)(3,6,7)(4,5,6)(4,5,7)(4,6,7)(5,6,7)
5 choose 4 = 5 combinations(1,2,3,4)(1,2,3,5)(1,2,4,5)(1,3,4,5)(2,3,4,5)
6 choose 4 = 15 combinations(1,2,3,4)(1,2,3,5)(1,2,3,6)(1,2,4,5)(1,2,4,6)(1,2,5,6)(1,3,4,5)(1,3,4,6)(1,3,5,6)(1,4,5,6)(2,3,4,5)(2,3,4,6)(2,3,5,6)(2,4,5,6)(3,4,5,6)
7 choose 4 = 35 combinations(1,2,3,4)(1,2,3,5)(1,2,3,6)(1,2,3,7)(1,2,4,5)(1,2,4,6)(1,2,4,7)(1,2,5,6)(1,2,5,7)(1,2,6,7)(1,3,4,5)(1,3,4,6)(1,3,4,7)(1,3,5,6)(1,3,5,7)(1,3,6,7)(1,4,5,6)(1,4,5,7)(1,4,6,7)(1,5,6,7)(2,3,4,5)(2,3,4,6)(2,3,4,7)(2,3,5,6)(2,3,5,7)(2,3,6,7)(2,4,5,6)(2,4,5,7)(2,4,6,7)(2,5,6,7)(3,4,5,6)(3,4,5,7)(3,4,6,7)(3,5,6,7)(4,5,6,7)
6 choose 5 = 6 combinations(1,2,3,4,5)(1,2,3,4,6)(1,2,3,5,6)(1,2,4,5,6)(1,3,4,5,6)(2,3,4,5,6)
7 choose 5 = 21 combinations(1,2,3,4,5)(1,2,3,4,6)(1,2,3,4,7)(1,2,3,5,6)(1,2,3,5,7)(1,2,3,6,7)(1,2,4,5,6)(1,2,4,5,7)(1,2,4,6,7)(1,2,5,6,7)(1,3,4,5,6)(1,3,4,5,7)(1,3,4,6,7)(1,3,5,6,7)(1,4,5,6,7)(2,3,4,5,6)(2,3,4,5,7)(2,3,4,6,7)(2,3,5,6,7)(2,4,5,6,7)(3,4,5,6,7)

How to take into account the order of the elements?

By principle, combinations do not take into account order (1,2) = (2,1). Use the function permutations to get ordered combinations.

How to get combinations with repetitions?

dCode has a dedicated tool for combinations with repetitions.

How many combinations is there to lottery/euromillions?

To win at EuroMillions, a player ticks 5 boxes out of 50 (50 choose 5), then 2 stars out of 11 (11 choose 2).

Example: Calculate the number of combinations of (50 choose 5) = 2 118 760, and multiply by (11 choose 2) = 55 for a total of 116 531 800 combinations.

To win at Powerball, pick 5 out of 69 (69 choose 5), then pick 1 out of 26 (26 choose 1).

Example: Calculate the number of combinations of (69 choose 5) = 11 238 513, and multiply by (26 choose 1) = 26 for a total of 292 201 338 combinations.

Many books describes strategies for lotto or lottery such as here. One of the strategies is to play covering designs systems.

Why k cannot be equal to zero 0?

If $ k = 0 $, then 0 item are wanted, there is an empty result with 0 item. So $$ \binom{n}{0} = 1 $$

Why n cannot be equal to zero 0?

If $ n = 0 $, then there is 0 item, impossible to pick $ k $, so there are no results. So $$ \binom{0}{k} = 0 $$

What is the value of 0 choose 0?

By convention $$ \binom{0}{0} = 1 $$

What is the algorithm for counting combinations?

// pseudo code
start count_combinations( k , n ) {
if (k = n) return 1;
if (k > n/2) k = n-k;
res = n-k+1;
for i = 2 by 1 while i < = k
res = res * (n-k+i)/i;
end for
return res;
end
// language C
double factorial(double x) {
double i;
double result=1;
if (x >= 0) {
for(i=x;i>1;i--) {
result = result*i;
}
return result;
}
return 0; // error
}
double count_combinations(double x,double y) {
double z = x-y;
return factorial(x)/(factorial(y)*factorial(z));
}

What is the algorithm to generate combinations?

// javascript
function combinations(a) { // a = new Array(1,2)
var fn = function(n, src, got, all) {
if (n == 0) {
if (got.length > 0) {
all[all.length] = got;
}
return;
}
for (var j = 0; j < src.length; j++) {
fn(n - 1, src.slice(j + 1), got.concat([src[j]]), all);
}
return;
}
var all = [];
for (var i=0; i < a.length; i++) {
fn(i, a, [], all);
}
all.push(a);
return all;
}

Source code

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Questions / Comments

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