Hay 12 estaciones intermedias entre dos lugares A y B. ¿Encuentre el número de maneras en que se puede hacer que un tren se detenga en 4 de estas estaciones intermedias para que no haya dos estaciones de parada consecutivas?
Ejemplos –
Input : n = 12, s = 4 Output : 126 Input : n = 16, s = 5 Output : 792
Explicación 1:
Fije/elimine las cuatro paradas como puntos fijos y calcule de cuántas maneras se pueden insertar las otras estaciones entre ellas, si debe tener al menos una estación entre paradas.
A x x x x x x x x B
Entre estas 8 estaciones sin paradas tenemos 9 lugares y seleccionamos estos 9 lugares como parada entre estas 8 estaciones.
Por lo tanto, la respuesta debería ser 
= 126
Explicación 2:
si conoce combinaciones de bolas indistinguibles en cajas distinguibles, simplemente puede usar, 
. En esta pregunta, $n$es el número de estaciones y $p$es el número de estaciones en las que desea detenerse. Aquí las estaciones de parada son como bolas indistinguibles y las estaciones sin parada son como cajas distinguibles.
Por lo tanto, 
= = 126
Código –
C++
#include <stdio.h>
int stopping_station(int, int);
// function to calculate number
// of ways of selecting 'p' non consecutive
// stations out of 'n' stations
int stopping_station(int p, int n)
{
int num = 1, dem = 1, s = p;
// selecting 's' positions out of 'n-s+1'
while (p != 1) {
dem *= p;
p--;
}
int t = n - s + 1;
while (t != (n - 2 * s + 1)) {
num *= t;
t--;
}
if ((n - s + 1) >= s)
printf("%d", num / dem);
else
// if conditions does not satisfy of combinatorics
printf("not possible");
}
// driver code
int main()
{
// n is total number of stations
// s is no. of stopping stations
int n, s;
// arguments of function are
// number of stopping station
// and total number of stations
stopping_station(4, 12);
}
Java
// Java code to calculate number
// of ways of selecting 'p' non
// consecutive stations out of
// 'n' stations
import java.io.*;
import java.util.*;
class GFG
{
public static int stopping_station(int p, int n)
{
int num = 1, dem = 1, s = p;
// selecting 's' positions out of 'n-s+1'
while (p != 1)
{
dem *= p;
p--;
}
int t = n - s + 1;
while (t != (n - 2 * s + 1))
{
num *= t;
t--;
}
if ((n - s + 1) >= s)
System.out.print(num / dem);
else
// if conditions does not satisfy of combinatorics
System.out.print("not possible");
return 0;
}
public static void main (String[] args)
{
// n is total number of stations
// s is no. of stopping stations
int n, s;
// arguments of function are
// number of stopping station
// and total number of stations
stopping_station(4, 12);
}
}
// ""This code is contributed by Mohit Gupta_OMG ""
Python3
# Python code to calculate number
# of ways of selecting 'p' non
# consecutive stations out of
# 'n' stations
def stopping_station( p, n):
num = 1
dem = 1
s = p
# selecting 's' positions
# out of 'n-s+1'
while p != 1:
dem *= p
p-=1
t = n - s + 1
while t != (n-2 * s + 1):
num *= t
t-=1
if (n - s + 1) >= s:
return int(num/dem)
else:
# if conditions does not
# satisfy of combinatorics
return -1
# driver code
num = stopping_station(4, 12)
if num != -1:
print(num)
else:
print("Not Possible")
# This code is contributed by "Abhishek Sharma 44"
C#
// C# code to calculate number
// of ways of selecting 'p' non
// consecutive stations out of
// 'n' stations
using System;
class GFG {
public static int stopping_station(int p,
int n)
{
int num = 1, dem = 1, s = p;
// selecting 's' positions
// out of 'n-s+1'
while (p != 1)
{
dem *= p;
p--;
}
int t = n - s + 1;
while (t != (n - 2 * s + 1))
{
num *= t;
t--;
}
if ((n - s + 1) >= s)
Console.WriteLine(num / dem);
// if conditions does not
// satisfy of combinatorics
else
Console.WriteLine("Not possible");
return 0;
}
// Driver Code
public static void Main(String []args)
{
// arguments of function are
// number of stopping station
// and total number of stations
stopping_station(4, 12);
}
}
// This code is contributed by vt_m.
PHP
<?php
// PHP code for Number of stopping
// station problem
// function to calculate number
// of ways of selecting 'p' non
// consecutive stations out of
// 'n' stations
function stopping_station(int $p, int $n)
{
$num = 1;
$dem = 1;
$s = $p;
// selecting 's' positions
// out of 'n-s+1'
while($p != 1)
{
$dem *= $p;
$p--;
}
$t = $n - $s + 1;
while($t != ($n - 2 * $s + 1))
{
$num *= $t;
$t--;
}
if (($n - $s + 1) >= $s)
echo $num / $dem;
else
// if conditions does not
// satisfy of combinatorics
echo "not possible";
}
// Driver Code
// n is total number of stations
// s is no. of stopping stations
$n; $s;
// arguments of function are
// number of stopping station
// and total number of stations
stopping_station(4, 12);
// This code is contributed by anuj_67.
?>
Javascript
<script>
// Javascript code to calculate number
// of ways of selecting 'p' non
// consecutive stations out of
// 'n' stations
function stopping_station( p, n)
{
var num = 1, dem = 1, s = p;
// selecting 's' positions
// out of 'n-s+1'
while (p != 1)
{
dem *= p;
p--;
}
var t = n - s + 1;
while (t != (n - 2 * s + 1))
{
num *= t;
t--;
}
if ((n - s + 1) >= s){
document.write(num / dem);
}
// if conditions does not
// satisfy of combinatorics
else{
document.write("Not possible");
}
return 0;
}
// Driver Code
// arguments of function are
// number of stopping station
// and total number of stations
stopping_station(4, 12);
</script>
Producción :
126
Publicación traducida automáticamente
Artículo escrito por Shivani Virmani y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA