이항계수 항등식으로 f(10) 구하기

$n \geq k \geq 3$일 때, $k ( k - 1 ) ( k - 2 ) _{n} {\mathrm{C}} _{k}$$\displaystyle = k ( k - 1 ) ( k - 2 ) \times \frac{n !}{k ! ( n - k ) !}$ $\displaystyle = \frac{n !}{( k - 3 ) ! ( n - k ) !}$ $\displaystyle = n ( n - 1 ) ( n - 2 ) \times \frac{( n - 3 ) !}{( k - 3 ) ! ( n - k ) !}$ $= n ( n - 1 ) ( n - 2 ) _{n-3} {\mathrm{C}} _{k-3}$ $\displaystyle \sum\limits _{k=0} ^{n} k ( k - 1 ) ( k - 2 ) _{n} {\mathrm{C}} _{k} p ^{k} ( 1 - p ) ^{n-k}$ $=$$\displaystyle \sum\limits _{k=3} ^{n} k ( k - 1 ) ( k - 2 ) _{n} {\mathrm{C}} _{k} p ^{k} ( 1 - p ) ^{n-k}$ $=$$\displaystyle \sum\limits _{k=3} ^{n} n ( n - 1 ) ( n - 2 ) _{n-3} {\mathrm{C}} _{k-3} p ^{k} ( 1 - p ) ^{n-k}$ $=$$\displaystyle n ( n - 1 ) ( n - 2 ) p ^{3} \sum\limits _{k=0} ^{n-3} {}_{n-3} {\mathrm{C}} _{k} p ^{k-3} ( 1 - p ) ^{n-3-k}$ $= n ( n - 1 ) ( n - 2 ) p ^{3}$$\left\{ p + ( 1 - p ) \right\} ^{n-3}$ $= n ( n - 1 ) ( n - 2 ) p ^{3}$ ∴ $f ( n ) = n ( n - 1 ) ( n - 2 )$ ∴ $f ( 10 ) = 10 \cdot 9 \cdot 8 = 720$