Solve it.....9

This is an easy problem. It is just $f\left( ab \right) \quad =\quad { ^{ a } }C_{ b }\quad if\quad a\quad >\quad b\\ and\quad f\left( ab \right) \quad =\quad { ^{ b } }C_{ a }\quad if\quad b\quad >\quad a\\$ $\quad \\ f(54)\quad =\quad { ^{ 5 } }C_{ 4 }\quad =\quad \frac { 5! }{ 4! } \quad =\quad 5\\ f(49)\quad =\quad { ^{ 9 } }C_{ 4 }\quad =\quad \frac { 9! }{ 5!\quad \times \quad 4! } \quad =\quad 126\\ f(57)\quad =\quad { ^{ 7 } }C_{ 5 }\quad =\quad \frac { 7! }{ 5!\quad \times \quad 2! } \quad =\quad 21\\ f(64)\quad =\quad { ^{ 6 } }C_{ 4 }\quad =\quad \frac { 6! }{ 4!\quad \times \quad 2! } \quad =\quad 15\\ f(98)\quad =\quad { ^{ 9 } }C_{ 8 }\quad =\quad \frac { 9! }{ 8! } \quad =\quad 9\\ So,\quad f(50)\quad =\quad { ^{ 5 } }C_{ 0 }\quad =\quad 0\quad as\quad \\ { ^{ n } }C_{ 0 }\quad =\quad 1$ Therefore 0 is the answer.

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