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

Answer: Total number of ways in which the seats can be distributed among parties such that any 2 of them combined and form a govt. is - 
C(2n+2, 2) - 3 C(n+2, 2) +3

I cannot understand why we have to add 3 with the final expression. This question is similar to this except the problem considering even number of seats.

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No. of ways to divide $2n$ seats among $3$ parties = no. of ways in which we can distribute $2n$ identical balls into $3$ distinct bins

$= {}^{2n+2}C_2.$

Now, we have to ensure that selecting any two party will get no. of seats more than the other. So, we can take the complement case where no party gets the majority of seats. Here, we have $2n$ seats and to ensure no majority one party needs $n$ seats. So, remaining $n$ seats we can distribute among $3$ parties in ${}^{n+2}C_2$ ways and the $n$ seats can go to any one among the $3$ party and thus we get $3 \times {}^{n+2}C_2$ ways. 

But we over counted above. This happens when $n$ seats go to say party A and $n$ remaining seats go to party B. This case will be repeated when initial $n$ seats go to party B and remaining $n$ to party A and similarly for B and C too. So, AB-BA, AC-CA, BC-CB, 3 cases are counted extra. Excluding these our final answer will be 

$ {}^{2n+2}C_2  - 3 \times {}^{n+2}C_2 + 3.$

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2N balls ...3 baskets .....distribute balls

O.....O|O......O|....

N           N     ...the third basket remained empty ....ways to do that (2N+2)!/2!2N!...ie 2N+2C2...

this was general distribution without condition ...now to follow condition ..i will give N balls already to any one basket ...

now i have N balls left ..that can be distributed in same way as did with 2N balls....N+2C2 *3 ..as i could put those first N balls in any basket ....now in this procedure ....i have counted ..O|O| ....twice ...how ..lets give N already to 1 basket ..and then to 2 ..and this will happen when N will be given already to 2nd basket first ..so need to add ...3 ..."HERE 3 are not special cases ...these are just overcounted ways ",,,,thats why its correct

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