Question

In how many ways can $2n+1$ seats in a congress be divided among 3 parties so that coalition of any 2 parties will ensure them  majority?

Answer 1

This problem corresponds to the problem of non negative integral solutions to the equation 

$P1+P2+P3= 2n+1$ where we have to distribute $2n+1$ seats(identical) among 3 parties(distinct) $P1,P2,P3$.

The solution will be ${}^{n-1+r}C_r$ having $n=3$ and $r=2n+1$. This comes as ${}^{ 2n+3}C_{2n+1}$  which further reduces to ${}^{2n+3}C_2=A$. (say). 

EDIT :  

The constraint the the coalition of 2 parties must form a govt can be dealt with as follows-

We have to also  ensure that govt must be formed by coalition so we have to eliminate the case where a single party gets a majority i.e. $n+1$ votes. That corresponds to non negative integral  solutions to the eqn 

$P1+P2+P3=n$  

solution will be ${}^3C_1 \times  {}^{n+2}C_n$  which reduces to  $3 \times {}^{n+2}C_2=B.$

Hence the final answer will be $A-B$ i.e.,    ${}^{ 2n+3}C_2 - 3 \times  {}^{n+2}C_2.$

 

Comments

when u will assign n+1 seats to any 1 team , then u will be left with n seats so the problem reduces to distributing n identical objects among 2 distinct parties rght ?

---

No, I think after assuring n+1 seats to one pary, we can still distribute the remaining n seats among all the 3 parties including the one which already got n+1 seats. Hence the equation P1+P2+P3= n

---

yes  after giving  ( N + 1) seats to one party distributing the remaining n seats among 3 parties