# In how many ways the ball will come back to the first boy after N turns

• Last Updated : 03 Mar, 2022

Four boys are playing a game with a ball. In each turn, the player (who has the ball currently) passes it to a different player randomly. Bob always starts the game. The task is to find in how many ways the ball will come back to Bob after N passes.
Examples:

Input: N = 3
Output:
Here are all the possible ways:
Bob -> Boy1 -> Boy2 -> Bob
Bob -> Boy1 -> Boy3 -> Bob
Bob -> Boy2 -> Boy1 -> Bob
Bob -> Boy2 -> Boy3 -> Bob
Bob -> Boy3 -> Boy1 -> Bob
Bob -> Boy3 -> Boy2 -> Bob
Input: N = 10
Output: 14763

Approach: Let the number of sequences that get back to Bob after N passes are P(N). The are two cases, either pass N – 2 is to Bob or it is not. Note that Bob can’t have the ball at (N – 1)th pass because then he won’t have the ball at the Nth pass.

1. Case 1: If pass N – 2 is to Bob then the pass N – 1 can be to any of the other 3 boys. Thus, the number of such sequences is 3 * P(N – 2).
2. Case 2: If pass N – 2 is not to Bob then pass N – 1 is to one of the 2 boys other than Bob and the one who got the ball in hand. So, substitute Bob for the receiver of pass N – 1, and get a unique N – 1 sequence. So, the number of such sequences are 2 * P(N – 1).

Hence the recurrence relation will be P(N) = 2 * P(N – 1) + 3 * P(N – 2) where P(0) = 1 and P(1) = 0.
After solving the recurrence relation, P(N) = (3N + 3 * (-1N)) / 4
Below is the implementation of the above approach:

## C++

 `// Function to return the number of``// sequences that get back to Bob``#include ``using` `namespace` `std;` `int` `numSeq(``int` `n)``{``    ``return` `(``pow``(3, n) + 3 * ``pow``(-1, n)) / 4;``}` `// Driver code``int` `main()``{``    ``int` `N = 10;``    ``printf``(``"%d"``, numSeq(N));``    ``return` `0;``}` `// This code is contributed by Mohit kumar`

## Java

 `// Function to return the number of``// sequences that get back to Bob``import` `java.util.*;` `class` `GFG``{` `static` `int` `numSeq(``int` `n)``{``    ``return` `(``int``) ((Math.pow(``3``, n) + ``3` `*``                    ``Math.pow(-``1``, n)) / ``4``);``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `N = ``10``;``    ``System.out.printf(``"%d"``, numSeq(N));``}``}` `// This code is contributed by Rajput-Ji`

## Python3

 `# Function to return the number of``# sequences that get back to Bob``def` `numSeq(n):``    ``return` `(``pow``(``3``, n) ``+` `3` `*` `pow``(``-``1``, n))``/``/``4``    ` `# Driver code``N ``=` `10``print``(numSeq(N))`

## C#

 `// C# implementation of the above approach``using` `System;` `// Function to return the number of``// sequences that get back to Bob``class` `GFG``{` `    ``static` `int` `numSeq(``int` `n)``    ``{``        ``return` `(``int``) ((Math.Pow(3, n) + 3 *``                       ``Math.Pow(-1, n)) / 4);``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int` `N = 10;``        ``Console.WriteLine(numSeq(N));``    ``}``}` `// This code is contributed by AnkitRai01`

## Javascript

 ``
Output:
`14763`

Time Complexity: O(log n)

Auxiliary Space: O(1)

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