Ahead of the pack

This is known as the ballot box problem .

With $A$ getting $p$ votes and $B$ getting $q$ votes, with $p \gt q$ , the general solution is

$\dbinom{p + q - 1}{p - 1} - \dbinom{p + q - 1}{p}.$

In this case we have $p = 11$ and $q = 7$ , and thus the solution is

$\dbinom{17}{10} - \dbinom{17}{11} = 7072.$

Proofs are provided in the link.

i do not know the solution through combinatorics. I solved it by writing recursive code for the counting process(it's in c).

#include <stdio.h>
int count=0;   //number of ways
const int Amax=11;
const int Bmax=7;      //final vote count
void vote(int a, int b){ //a & b are their current vote counts
    if(a==Amax||b==Bmax){ //Base case
        count++;
        return;
    }
    vote(a+1, b);     //let one vote for A and we proceed counting
    if(b<a-1){
        vote(a, b+1);    //let one vote for B & counting ahead
    }
}
int main() {
    vote(0, 0); 
    printf("%d", count);
    return 0;
}

you can run this code for different values of votes.

just change Amax & Bmax in lines #3 & #4

remember Amax>Bmax

I do not know how to solve this problem yet. It appeared in an exam that I took. In the solutions, only the following formula was given:

$\begin{pmatrix} m+n \\ n \end{pmatrix}\times \frac { (m-n) }{ (m+n) } =7072$

Where, $m=11, n=7$ and $\begin{pmatrix} N \\ r \end{pmatrix}$ is the coefficient of $x^r$ in $(1+x)^N$ .

Please post ideas/solutions. Thanks!

I didn't know about ballot box problem, so I solved using my own technique as follows

Where $n_1\ n_2$ denotes A has got $n_1$ votes while B has got $n_2$ votes. The values in () are the number of ways to get to $n_1\ n_2$ . The idea is similar to pascal's triangle with the restrictions.