Answer:
/* Program in Java for finding possible ways of arriving exactly at the finishing point as mentioned above*/
import java.io.*;
import java.util.Scanner;
class Main
{
public static int fact(int num)
{
int i,m,fact=1;
int number;
System.out.println("Enter n:");
Scanner sm= new Scanner(System.in);
number= sm.nextInt();
for(i=1;i<=number;i++)
{
fact=fact*i;
}
return fact;
}
public static void main(String args[])
{
int m;
System.out.println("Enter the number of times dice is rolled:");
Scanner sn = new Scanner(System.in);
m= sn.nextInt();
int numofpossibleways= 126*fact(m)/fact(m-6);
System.out.print("Number of possible ways:"+ numofpossibleways);
}
}
Explanation:
Suppose
starting point=0
spaces traveled=n
suppose there are m number of times dice is rolled.
hence for 6,5,4,3,2,1 it will be m(m-1)(m-2)(m-3)(m-4)(m-5)
So total ways to reach n spaces away from starting point:
(6* 6C1 +5*6C1+ 4*6C1+3*6C1+2*6C1+!*6C1)m!/(m-6)!
=((6+5+4+3+2+1)*6C1))m!/(m-6)!=(21*6!/1!5!)*m!/(m-6!)
=21*6= 126m!/'(m-6)!
