In how many ways “ n ” toys can be distributed among 3 friends so that there is not more than one winner and all of them contain at least one toy?

Question

In how many ways "n" toys can be distributed among 3 friends so that there is not more than one winner and all of them contain at least one toy?

Explanation:

For n=5 , We can distribute in the following 3 ways:

(1,1,3), (1,3,1), (3,1,1) .

Answer 1

Each player gets N/3 of them.

Then you find the REMAINDER (using the mod function) and add it to one of them,

so that is the unique winner

So for 5 toys, 5/3 = 1, so each player gets one toy.

5 mod 3 = 2, so given the 2 toys to one player.

Therefore, (1,3,1), (3,1,1) or (1,1,3)

For n=7, 7/3 = 2 so each players gets 2 toys.

7 mod 3 = 1, so give the 1 extra toy to one player.

(2,3,2) (3,2,2) or (2,2,3)

Now, IF the number N is evenly divisible by 3, then

just give an extra toy to another player.

EX. N=9. Each player gets 9/3=3 toys with none

left over. So give one toy another player.

Then (4,3,2), (3,4,2), (2,3,4)

In this case, the solution shall be

(k+1,k,k-1) where k = N/3

Here's Java:

import java.io.*;

class ToyDistribution

{

private Console console;

private int numToys;

ToyDistribution()

{

console = System.console();

}

public void Go()

{

try

{

System.out.print(" how many toys ??? :> ");

String inbuff = console.readLine();

numToys = Integer.parseInt(inbuff);

int remainder = numToys % 3;

int quotient = numToys / 3;

if (remainder==0)

{

System.out.println( "(" + (quotient+1) + " , "

+ (quotient) + " , "

+ (quotient-1) + ")"

);

}

else

{

System.out.println(

"(" + (quotient) + " , "

+ (quotient + remainder) + " , "

+ (quotient) + ")"

);

}

}

catch (Exception ex)

{

System.out.println(" Error on input - sorry ");

}

}

public static void main(String args[])

{

ToyDistribution x = new ToyDistribution();

x.Go();

}

}