Firemoon and Watersun Tournament

$2^{2n}$ = total number of ways.

$^{ 2n }{ C }_{ n }$ = number of ways in which each of them wins n matches.

as series has to be won so total ways in which series can be won is

$2^{ 2n }-^{ 2n }{ C }_{ n }$

no. of ways series can be won is same for both so for one ofthem

= $\frac { 2^{ 2n }-^{ 2n }{ C }_{ n } }{ 2 }$

put n=14 and get the answer

Total number of matches = $2n$

To win the majority, WATERSUN (the girl :P) must win at least $n+1$ matches. So, the ways in which she can do this is :

${ 2n }_{ { C }_{ n+1 } }+{ 2n }_{ { C }_{ n+2 } }+2n_{ { C }_{ n+3 } }+......{ 2n }_{ { C }_{ 2n-1 } }+2n_{ { C }_{ 2n } }$ , where,

${ a }_{ { C }_{ b } }$ , signifies the number of ways to choose $b$ objects out of $a$ distinct objects. So, the required term means number of ways of selecting more than half matches to win. Now, to solve this HUGE term:

We know that:

${ (1+x) }^{ 2n } = { 2n }_{ { C }_{ 0 } }{ x }^{ 2n }+{ 2n }_{ { C }_{ 1 } }{ x }^{ 2n-1 }+......+2n_{ { C }_{ n } }{ x }^{ n }+......{ 2n }_{ { C }_{ 2n-1 } }{ x }^{ 1 }+2n_{ { C }_{ 2n } }{ x }^{ 0 }$

But,

${ 2n }_{ { C }_{ 0 } }={ 2n }_{ { C }_{ 2n } }, { 2n }_{ { C }_{ 1 } }={ 2n }_{ { C }_{ 2n-1 } } ..... { 2n }_{ { C }_{ n-1 } }={ 2n }_{ { C }_{ n+1 } }$ --------------1.

So, our base equation becomes (USING 1.):

${ (1+x) }^{ 2n }-2n_{ { C }_{ n } }{ x }^{ n }\quad =\quad { 2n }_{ { C }_{ 0 } }{ x }^{ 2n }+{ 2n }_{ { C }_{ 1 } }{ x }^{ 2n-1 }+......+{ 2n }_{ { C }_{ 2n-1 } }{ x }^{ 1 }+2n_{ { C }_{ 2n } }{ x }^{ 0 }\\ \Rightarrow { (1+x) }^{ 2n }-2n_{ { C }_{ n } }{ x }^{ n } = { 2n }_{ { C }_{ 2n } }{ x }^{ 2n }+{ 2n }_{ { C }_{ 2n-1 } }{ x }^{ 2n-1 }+......+{ 2n }_{ { C }_{ 2n-1 } }{ x }^{ 1 }+2n_{ { C }_{ 2n } }{ x }^{ 0 }$

Put x=1:

$\Rightarrow { 2 }^{ 2n }-2n_{ { C }_{ n } }{ 1 }^{ n }\quad =\quad { 2n }_{ { C }_{ 0 } }{ 1 }^{ 2n }+{ 2n }_{ { C }_{ 1 } }{ 1 }^{ 2n-1 }+......+{ 2n }_{ { C }_{ 2n-1 } }{ 1 }^{ 1 }+2n_{ { C }_{ 2n } }{ 1 }^{ 0 }\\ \Rightarrow 2^{ 2n }-2n_{ { C }_{ n } }\quad =\quad { 2n }_{ { C }_{ 2n } }+{ 2n }_{ { C }_{ 2n-1 } }+...{ 2n }_{ { C }_{ n+1 } }+{ 2n }_{ { C }_{ n+1 } }...+{ 2n }_{ { C }_{ 2n-1 } }+2n_{ { C }_{ 2n } }\\ \Rightarrow 2^{ 2n }-2n_{ { C }_{ n } }\quad =\quad 2({ 2n }_{ { C }_{ n+1 } }+{ 2n }_{ { C }_{ n+2 } }+.....{ 2n }_{ { C }_{ 2n-1 } }+{ 2n }_{ { C }_{ 2n } })\\ \Rightarrow { 2n }_{ { C }_{ n+1 } }+{ 2n }_{ { C }_{ n+2 } }+.....{ 2n }_{ { C }_{ 2n-1 } }+{ 2n }_{ { C }_{ 2n } }\quad =\quad \frac { 2^{ 2n }-2n_{ { C }_{ n } } }{ 2 }$

And VOLA!! The term on the left hand side is our required expression. Now, simply putting the value of $n = 14$ , and using a bit of calculator, we get the value of:

${ 28 }_{ { C }_{ 15 } }+{ 28 }_{ { C }_{ 16 } }+.....{ 28 }_{ { C }_{ 27 } }+{ 28 }_{ { C }_{ 28 } }\quad =\quad \frac { 2^{ 28 }-28_{ { C }_{ 14 } } }{ 2 } \quad =\quad 114159428$

Cheers!

Well, Firemoon is a famous indian actor: Jeetendra, and his lover's name is Leena Chandavarkar. But no ideas, on the reason! Lol :D:D Good luck all...Hehe