Hướng dẫn nchoosek python

Here is a more complete answer. You should use itertools.combinations and not the itertools.permutations as combination is very different from permutation.

For instance if you needed all the two element combinations of an array such as [1,2,3,5] the following code will produce the result that you want (equivalent of nchoosek in Matlab)

>>> import itertools
>>> all_combos = list(itertools.combinations([1,2,3,5], 2))
>>> print all_combos
[(1, 2), (1, 3), (1, 5), (2, 3), (2, 5), (3, 5)]

if You would like all the combinations as a 2d array just convert the tuple list to a numpy array using the following command:

>>> all_combos = np.array(list(itertools.combinations([1,2,3,5], 2)))
>>> print all_combos
[[1 2]
[1 3]
[1 5]
[2 3]
[2 5]
[3 5]]

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(defn binom [n k]
  (let [fact #(apply * (range 1 (inc %)))]
    (/ (fact n)
       (* (fact k) (fact (- n k))))))

public BigInteger binom(int n, int k)
{
	return factorial(n)/(factorial(k) * factorial(n-k));
}

public BigInteger factorial(int x)
{
	BigInteger result = 1;
	for(int i=1;i<=x;i++)
	{
	result = result * i;
	}
	return result;
}

BigInt binom(uint n, uint k)
{
   assert(n >= k);
   BigInt r = 1;
   for(uint i = 0; i <= k; ++i)
   {
      r *= n-i;
      r /= i+1;
   }
   return r;
}

int binom(int n, int k) {
  int result = 1;
  for (int i = 0; i < k; i++) {
    result = result * (n - i) ~/ (i + 1);
  }
  return result;
}

integer, parameter :: i8 = selected_int_kind(18)
integer, parameter :: dp = selected_real_kind(15)
n = 100
k = 5
print *,nint(exp(log_gamma(n+1.0_dp)-log_gamma(n-k+1.0_dp)-log_gamma(k+1.0_dp)),kind=i8)

z := new(big.Int)
z.Binomial(n, k)

• Demo
• Doc

binom n k = product [1+n-k..n] `div` product [1..k]

const fac = x => x ? x * fac (x - 1) : x + 1
const binom = (n, k) => fac (n) / fac (k) / fac (n - k >= 0 ? n - k : NaN)

import java.math.BigInteger;

static BigInteger binom(int N, int K) {
    BigInteger ret = BigInteger.ONE;
    for (int k = 0; k < K; k++) {
        ret = ret.multiply(BigInteger.valueOf(N-k))
                 .divide(BigInteger.valueOf(k+1));
    }
    return ret;
}

• Demo
• Origin

• Demo
• Doc

var
  N, K, Res: TValue;
begin
  Res := BinomN(N, K);
end.

sub binom {
  my ($n, $k) = @_;
  if ($k > $n - $k) { $k = $n - $k }
  my $r = 1;
  for ( my $i = $n/$n ; $i <= $k;) {
    $r *= $n-- / $i++
  }
  return $r
}

sub binom {
   my ($n, $k) = @_;
   my $fact = sub {
      my $n = shift;
      return $n<2 ? 1 : $n * $fact->($n-1);
   };
   return $fact->($n) / ($fact->($k) * ($fact->($n-$k)));
}

def binom(n,k)
  (1+n-k..n).inject(:*)/(1..k).inject(:*)
end

extern crate num;

use num::bigint::BigInt;
use num::bigint::ToBigInt;
use num::traits::One;

fn binom(n: u64, k: u64) -> BigInt {
    let mut res = BigInt::one();
    for i in 0..k {
        res = (res * (n - i).to_bigint().unwrap()) /
              (i + 1).to_bigint().unwrap();
    }
    res
}

• Demo

• Python

def binom(n, k):
    return math.factorial(n) // math.factorial(k) // math.factorial(n - k)

• Python

def binom(n, k):
    return math.comb(n, k)

• Clojure

(defn binom [n k]
  (let [fact #(apply * (range 1 (inc %)))]
    (/ (fact n)
       (* (fact k) (fact (- n k))))))

• C#

public BigInteger binom(int n, int k)
{
	return factorial(n)/(factorial(k) * factorial(n-k));
}

public BigInteger factorial(int x)
{
	BigInteger result = 1;
	for(int i=1;i<=x;i++)
	{
	result = result * i;
	}
	return result;
}

• D

BigInt binom(uint n, uint k)
{
   assert(n >= k);
   BigInt r = 1;
   for(uint i = 0; i <= k; ++i)
   {
      r *= n-i;
      r /= i+1;
   }
   return r;
}

• Dart

int binom(int n, int k) {
  int result = 1;
  for (int i = 0; i < k; i++) {
    result = result * (n - i) ~/ (i + 1);
  }
  return result;
}

• Fortran

integer, parameter :: i8 = selected_int_kind(18)
integer, parameter :: dp = selected_real_kind(15)
n = 100
k = 5
print *,nint(exp(log_gamma(n+1.0_dp)-log_gamma(n-k+1.0_dp)-log_gamma(k+1.0_dp)),kind=i8)

• Go

z := new(big.Int)
z.Binomial(n, k)

• Haskell

binom n k = product [1+n-k..n] `div` product [1..k]

• JS

const fac = x => x ? x * fac (x - 1) : x + 1
const binom = (n, k) => fac (n) / fac (k) / fac (n - k >= 0 ? n - k : NaN)

• Java

static BigInteger binom(int N, int K) {
    BigInteger ret = BigInteger.ONE;
    for (int k = 0; k < K; k++) {
        ret = ret.multiply(BigInteger.valueOf(N-k))
                 .divide(BigInteger.valueOf(k+1));
    }
    return ret;
}

• PHP

gmp_binomial($n, $k);

• Pascal

var
  N, K, Res: TValue;
begin
  Res := BinomN(N, K);
end.

• Perl

sub binom {
  my ($n, $k) = @_;
  if ($k > $n - $k) { $k = $n - $k }
  my $r = 1;
  for ( my $i = $n/$n ; $i <= $k;) {
    $r *= $n-- / $i++
  }
  return $r
}

• Perl

sub binom {
   my ($n, $k) = @_;
   my $fact = sub {
      my $n = shift;
      return $n<2 ? 1 : $n * $fact->($n-1);
   };
   return $fact->($n) / ($fact->($k) * ($fact->($n-$k)));
}

• Ruby

def binom(n,k)
  (1+n-k..n).inject(:*)/(1..k).inject(:*)
end

• Rust

fn binom(n: u64, k: u64) -> BigInt {
    let mut res = BigInt::one();
    for i in 0..k {
        res = (res * (n - i).to_bigint().unwrap()) /
              (i + 1).to_bigint().unwrap();
    }
    res
}