There are n people numbered from 1
to n
. I have to write a code which produces and print all different combinations of k
people from these n
. Please explain the algorithm used for that.
To calculate combinations, we will use the formula nCr = n! / r! * (n - r)!, where n represents the total number of items, and r represents the number of items being chosen at a time. To calculate a combination, you will need to calculate a factorial.
Add a Custom Column to and name it List1. Enter the formula =List1. Expand out the new List1 column and then Close & Load the query to a table. The table will have all the combinations of items from both lists and we saved on making a custom column in List1 and avoided using a merge query altogether!
The number of combinations of n distinct objects, taken r at a time is: Cr = n! / r! (n - r)! Thus, 27,405 different groupings of 4 players are possible.
I assume you're asking about combinations in combinatorial sense (that is, order of elements doesn't matter, so [1 2 3]
is the same as [2 1 3]
). The idea is pretty simple then, if you understand induction / recursion: to get all K
-element combinations, you first pick initial element of a combination out of existing set of people, and then you "concatenate" this initial element with all possible combinations of K-1
people produced from elements that succeed the initial element.
As an example, let's say we want to take all combinations of 3 people from a set of 5 people. Then all possible combinations of 3 people can be expressed in terms of all possible combinations of 2 people:
comb({ 1 2 3 4 5 }, 3) = { 1, comb({ 2 3 4 5 }, 2) } and { 2, comb({ 3 4 5 }, 2) } and { 3, comb({ 4 5 }, 2) }
Here's C++ code that implements this idea:
#include <iostream> #include <vector> using namespace std; vector<int> people; vector<int> combination; void pretty_print(const vector<int>& v) { static int count = 0; cout << "combination no " << (++count) << ": [ "; for (int i = 0; i < v.size(); ++i) { cout << v[i] << " "; } cout << "] " << endl; } void go(int offset, int k) { if (k == 0) { pretty_print(combination); return; } for (int i = offset; i <= people.size() - k; ++i) { combination.push_back(people[i]); go(i+1, k-1); combination.pop_back(); } } int main() { int n = 5, k = 3; for (int i = 0; i < n; ++i) { people.push_back(i+1); } go(0, k); return 0; }
And here's output for N = 5, K = 3
:
combination no 1: [ 1 2 3 ] combination no 2: [ 1 2 4 ] combination no 3: [ 1 2 5 ] combination no 4: [ 1 3 4 ] combination no 5: [ 1 3 5 ] combination no 6: [ 1 4 5 ] combination no 7: [ 2 3 4 ] combination no 8: [ 2 3 5 ] combination no 9: [ 2 4 5 ] combination no 10: [ 3 4 5 ]
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