I have 10 words. How can I get all possible combinations of 5 words (n=10, k=5)
. The order does not matter.
For example: "A", "B", "C", if k=2 (n=3 in this case)
, it would like AB, BC and AC. Maybe you know some usefull code or example.
P.S. Sorry if I'm not right enough cause I don't know English very good.
Combinations are selections of objects, with or without repetition, order does not matter. The number of k-element combinations of n objects, without repetition is Cn,k = (n k ) = n! k!( n − k)! .
Answer and Explanation: The number of possible combinations with 4 numbers without repetition is 15. The formula we use to calculate the number of n element combinations when repetition is not allowed is 2n - 1.
Note that your choice of 5 objects can take any order whatsoever, because your choice each time can be any of the remaining objects. So we say that there are 5 factorial = 5! = 5x4x3x2x1 = 120 ways to arrange five objects.
Answer and Explanation: If repetition is allowed, then the number of permutations of 10 digits is 10,000,000,000. If repetition is not allowed, then the number of permutations of 10 digits is 3,628,800.
What you are trying to do is get all the permutations of a collection.
Here is the code snippet:
static void Main(string[] args)
{
var list = new List<string> { "a", "b", "c", "d", "e" };
var result = GetPermutations(list, 3);
foreach (var perm in result)
{
foreach (var c in perm)
{
Console.Write(c + " ");
}
Console.WriteLine();
}
Console.ReadKey();
}
static IEnumerable<IEnumerable<T>> GetPermutations<T>(IEnumerable<T> items, int count)
{
int i = 0;
foreach (var item in items)
{
if (count == 1)
yield return new T[] { item };
else
{
foreach (var result in GetPermutations(items.Skip(i + 1), count - 1))
yield return new T[] { item }.Concat(result);
}
++i;
}
}
Outputs:
a b c
a b d
a b e
a c d
a c e
a d e
b c d
b c e
b d e
c d e
Here's what I put together:
static class LinqExtensions
{
public static IEnumerable<IEnumerable<T>> CombinationsWithoutRepetition<T>(
this IEnumerable<T> items,
int ofLength)
{
return (ofLength == 1) ?
items.Select(item => new[] { item }) :
items.SelectMany((item, i) => items.Skip(i + 1)
.CombinationsWithoutRepetition(ofLength - 1)
.Select(result => new T[] { item }.Concat(result)));
}
public static IEnumerable<IEnumerable<T>> CombinationsWithoutRepetition<T>(
this IEnumerable<T> items,
int ofLength,
int upToLength)
{
return Enumerable.Range(ofLength, Math.Max(0, upToLength - ofLength + 1))
.SelectMany(len => items.CombinationsWithoutRepetition(ofLength: len));
}
}
...
foreach (var c in new[] {"a","b","c","d"}.CombinationsWithoutRepetition(ofLength: 2, upToLength: 4))
{
Console.WriteLine(string.Join(',', c));
}
produces:
a,b
a,c
a,d
b,c
b,d
c,d
a,b,c
a,b,d
a,c,d
b,c,d
a,b,c,d
Note that this is concise but inefficient and should not be used for large sets or inner loops. Notably, the short arrays are re-created multiple times and could be memoized, and the IEnumerable
will be iterated multiple times, which can cause unexpected work if care is not taken.
Also, if the input contains duplicates then the output will as well. Either use .Distinct().ToArray()
first, or use another solution which includes equality checking and, presumably, takes an IEqualityComparer
for generality.
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