一次排列N个对象的K个不同对象的排列方式数目

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📜 一次排列N个对象的K个不同对象的排列方式数目

📅 最后修改于: 2021-06-25 13:28:57 🧑 作者: Mango

给定两个整数K和N ，任务是计算一次放置N个对象的K个不同对象的排列方式的数量。每个布置最多包含一个对象。答案可能非常大，因此请以10 ⁹ + 7为模返回答案。

注意： 1 <= N <= K <= 10 ⁵ 。
先决条件：阶乘，计算nCr％p

例子：

Input : N = 3, K = 3
Output : 6
If 1, 2 and 3 be the K different objects, then the possible arrangements are {123, 132, 213, 231, 312, 321}.

Input : N = 4, K = 6
Output : 360

为什么编程需要懂一点英语

方法：可以使用置换和组合解决问题。由于N总是小于或等于K ，所以答案总是大于0。在一个排列中， K个可用对象中的任何对象最多可以使用一次。因此，我们需要为单个排列从K个可用对象中选择N个对象。因此，从K个对象中选择N个对象的方法总数为^K C _N。然后，可以以N种方式在它们之间对N个对象的任何选定组进行置换。因此，问题的答案是：

(N! * ^KC_N) % (10⁹ +7)

为什么编程需要懂一点英语

下面是上述方法的实现：

C++

// C++ implementation of the approach
  
#include 
using namespace std;
#define ll long long
#define mod (ll)(1e9 + 7)
  
// Function to return n! % p
ll factorial(ll n, ll p)
{
    ll res = 1; // Initialize result
    for (int i = 2; i <= n; i++)
        res = (res * i) % p;
    return res;
}
  
// Iterative Function to calculate (x^y)%p
// in O(log y)
ll power(ll x, ll y, ll p)
{
    ll res = 1; // Initialize result
  
    x = x % p; // Update x if it is more than or
    // equal to p
  
    while (y > 0) {
        // If y is odd, multiply x with result
        if (y & 1)
            res = (res * x) % p;
  
        // y must be even now
        y = y >> 1; // y = y/2
        x = (x * x) % p;
    }
    return res;
}
  
// Returns n^(-1) mod p
ll modInverse(ll n, ll p)
{
    return power(n, p - 2, p);
}
  
// Returns nCr % p using Fermat's little
// theorem.
ll nCrModP(ll n, ll r, ll p)
{
    // Base case
    if (r == 0)
        return 1;
  
    // Fill factorial array so that we
    // can find all factorial of r, n
    // and n-r
    ll fac[n + 1];
    fac[0] = 1;
    for (int i = 1; i <= n; i++)
        fac[i] = fac[i - 1] * i % p;
  
    return (fac[n] * modInverse(fac[r], p) % p 
                    * modInverse(fac[n - r], p) % p) % p;
}
  
// Function to return the number of ways to
// arrange K different objects taking N objects
// at a time
ll countArrangements(ll n, ll k, ll p)
{
    return (factorial(n, p) * nCrModP(k, n, p)) % p;
}
  
// Drivers Code
int main()
{
    ll N = 5, K = 8;
  
    // Function call
    cout << countArrangements(N, K, mod);
  
    return 0;
}

Java

// Java implementation of the approach
  
class GFG
{
      
static long mod =(long) (1e9 + 7);
  
// Function to return n! % p
static long factorial(long n, long p)
{
    long res = 1; // Initialize result
    for (int i = 2; i <= n; i++)
        res = (res * i) % p;
    return res;
}
  
// Iterative Function to calculate (x^y)%p
// in O(log y)
static long power(long x, long y, long p)
{
    long res = 1; // Initialize result
  
    x = x % p; // Update x if it is more than or
    // equal to p
  
    while (y > 0) 
    {
        // If y is odd, multiply x with result
        if ((y & 1)==1)
            res = (res * x) % p;
  
        // y must be even now
        y = y >> 1; // y = y/2
        x = (x * x) % p;
    }
    return res;
}
  
// Returns n^(-1) mod p
static long modInverse(long n, long p)
{
    return power(n, p - 2, p);
}
  
// Returns nCr % p using Fermat's little
// theorem.
static long nCrModP(long n, long r, long p)
{
    // Base case
    if (r == 0)
        return 1;
  
    // Fill factorial array so that we
    // can find all factorial of r, n
    // and n-r
    long fac[] = new long[(int)n + 1];
    fac[0] = 1;
    for (int i = 1; i <= n; i++)
        fac[i] = fac[i - 1] * i % p;
  
    return (fac[(int)n] * modInverse(fac[(int)r], p) % p 
                    * modInverse(fac[(int)n - (int)r], p) % p) % p;
}
  
// Function to return the number of ways to
// arrange K different objects taking N objects
// at a time
static long countArrangements(long n, long k, long p)
{
    return (factorial(n, p) * nCrModP(k, n, p)) % p;
}
  
// Driver Code
public static void main(String[] args)
{
    long N = 5, K = 8;
    // Function call
    System.out.println(countArrangements(N, K, mod));
}
}
  
/* This code is contributed by PrinciRaj1992 */

Python3

# Python3 implementation of the approach
mod = 10**9 + 7
  
# Function to return n! % p
def factorial(n, p):
  
    res = 1 # Initialize result
    for i in range(2, n + 1):
        res = (res * i) % p
    return res
  
# Iterative Function to calculate
# (x^y)%p in O(log y)
def power(x, y, p):
  
    res = 1 # Initialize result
  
    x = x % p # Update x if it is 
              # more than or equal to p
  
    while (y > 0):
          
        # If y is odd,
        # multiply x with result
        if (y & 1):
            res = (res * x) % p
  
        # y must be even now
        y = y >> 1 # y = y/2
        x = (x * x) % p
      
    return res
  
# Returns n^(-1) mod p
def modInverse(n, p):
  
    return power(n, p - 2, p)
  
# Returns nCr % p using Fermat's little
# theorem.
def nCrModP(n, r, p):
  
    # Base case
    if (r == 0):
        return 1
  
    # Fifactorial array so that we
    # can find afactorial of r, n
    # and n-r
    fac = [0 for i in range(n + 1)]
    fac[0] = 1
    for i in range(1, n + 1):
        fac[i] = fac[i - 1] * i % p
  
    return (fac[n] * modInverse(fac[r], p) % p * 
                     modInverse(fac[n - r], p) % p) % p
  
# Function to return the number of ways 
# to arrange K different objects taking 
# N objects at a time
def countArrangements(n, k, p):
  
    return (factorial(n, p) * 
            nCrModP(k, n, p)) % p
  
# Drivers Code
N = 5
K = 8
  
# Function call
print(countArrangements(N, K, mod))
  
# This code is contributed by mohit kumar

C#

// C# implementation of the approach
using System;
  
class GFG
{
      
static long mod =(long) (1e9 + 7);
  
// Function to return n! % p
static long factorial(long n, long p)
{
    long res = 1; // Initialize result
    for (int i = 2; i <= n; i++)
        res = (res * i) % p;
    return res;
}
  
// Iterative Function to calculate (x^y)%p
// in O(log y)
static long power(long x, long y, long p)
{
    long res = 1; // Initialize result
  
    x = x % p; // Update x if it is more than or
    // equal to p
  
    while (y > 0) 
    {
        // If y is odd, multiply x with result
        if ((y & 1)==1)
            res = (res * x) % p;
  
        // y must be even now
        y = y >> 1; // y = y/2
        x = (x * x) % p;
    }
    return res;
}
  
// Returns n^(-1) mod p
static long modInverse(long n, long p)
{
    return power(n, p - 2, p);
}
  
// Returns nCr % p using Fermat's little
// theorem.
static long nCrModP(long n, long r, long p)
{
    // Base case
    if (r == 0)
        return 1;
  
    // Fill factorial array so that we
    // can find all factorial of r, n
    // and n-r
    long[] fac = new long[(int)n + 1];
    fac[0] = 1;
    for (int i = 1; i <= n; i++)
        fac[i] = fac[i - 1] * i % p;
  
    return (fac[(int)n] * modInverse(fac[(int)r], p) % p 
                    * modInverse(fac[(int)n - (int)r], p) % p) % p;
}
  
// Function to return the number of ways to
// arrange K different objects taking N objects
// at a time
static long countArrangements(long n, long k, long p)
{
    return (factorial(n, p) * nCrModP(k, n, p)) % p;
}
  
// Driver Code
public static void Main()
{
    long N = 5, K = 8;
      
    // Function call
    Console.WriteLine(countArrangements(N, K, mod));
}
}
  
/* This code is contributed by Code_Mech */

PHP

 0) 
    {
        // If y is odd, multiply x with result
        if (($y & 1)==1)
            $res = ($res * $x) % $p;
  
        // y must be even now
        $y = $y >> 1; // y = y/2
        $x = ($x * $x) % $p;
    }
    return $res;
}
  
// Returns n^(-1) mod p
function modInverse($n, $p)
{
    return power($n, $p - 2, $p);
}
  
// Returns nCr % p using Fermat's little
// theorem.
function nCrModP($n, $r, $p)
{
    // Base case
    if ($r == 0)
        return 1;
  
    // Fill factorial array so that we
    // can find all factorial of r, n
    // and n-r
    $fac= array((int)$n + 1);
    $fac[0] = 1;
    for ($i = 1; $i <= $n; $i++)
        $fac[$i] = $fac[$i - 1] * $i % $p;
  
    return ($fac[(int)$n] * modInverse($fac[(int)$r], $p) % $p 
                    * modInverse($fac[(int)$n - (int)$r], $p) % $p) % $p;
}
  
// Function to return the number of ways to
// arrange K different objects taking N objects
// at a time
function countArrangements($n, $k, $p)
{
    return (factorial($n, $p) * nCrModP($k, $n, $p)) % $p;
}
  
// Driver Code
{
    $N = 5; $K = 8;
    // Function call
    echo(countArrangements($N, $K, $mod));
}
  
  
/* This code is contributed by Code_Mech*/

输出：

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