Số cách là :
$C_n^0(C_n^1.3^{n-1}+C_n^2.3^{n-2}+C_n^3.3^{n-3}+C_n^4.3^{n-4}+C_n^5.3^{n-5})+C_n^1(C_{n-1}^2.3^{n-3}+C_{n-1}^3.3^{n-4}+C_{n-1}^4.3^{n-5})+C_n^2C_{n-2}^3.3^{n-5}=C_n^1.3^{n-1}+C_n^2.3^{n-2}+(C_n^0C_n^3+C_n^1C_{n-1}^2).3^{n-3}+(C_n^0C_n^4+C_n^1C_{n-1}^3).3^{n-4}+(C_n^0C_n^5+C_n^1C_{n-1}^4+C_n^2C_{n-2}^3).3^{n-5}$
$=C_n^1.3^{n-1}+C_n^2.3^{n-2}+4C_n^3.3^{n-3}+5C_n^4.3^{n-4}+16C_n^5.3^{n-5}$.
Sao anh không dùng hàm sinh!
Ta có hàm sinh :
$\begin{align*}
f(x) &=e^{3x}\left [\left (x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!} \right )+x\left (\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}\right )+\frac {x^2}{2!}.\left (\frac{x^3}{3!} \right )\right ]\\
&=e^{3x}\left ( x+\frac {x^2}{2!}+4.\frac{x^3}{3!}+5.\frac{x^4}{4!}+16.\frac{x^5}{5!} \right )\\
\Longrightarrow \left [ x^n \right ]f(x)&=\left [ x^{n-1} \right ]e^{3x}+\left [ x^{n-2} \right ]\frac {e^{3x}}{2!}+4.\left [ x^{n-3} \right ]\frac {e^{3x}}{3!}+5.\left [ x^{n-4} \right ]\frac {e^{3x}}{4!}+16.\left [ x^{n-5} \right ]\frac {e^{3x}}{5!}\\
\Longrightarrow n!\left [ x^n \right ]f(x)&=n3^{n-1}+\frac{1}{2}n(n-1)3^{n-2}+\frac{2}{3}n(n-1)(n-2)3^{n-3}+\frac{5}{24}n(n-1)(n-2)(n-3)3^{n-4}\\
&+\frac{2}{15}n(n-1)(n-2)(n-3)(n-4)3^{n-5}
\end{align*}$
Gọi $a_n$ là số cách phát n viên bi thì:
$\boldsymbol {a_n=\begin {cases}
1&n=1,\\
n3^{n-1}+\frac{1}{2}n(n-1)3^{n-2}&n=2,\\
n3^{n-1}+\frac{1}{2}n(n-1)3^{n-2}+\frac{2}{3}n(n-1)(n-2)3^{n-3}&n=3,\\
n3^{n-1}+\frac{1}{2}n(n-1)3^{n-2}+\frac{2}{3}n(n-1)(n-2)3^{n-3}+\frac{5}{24}n(n-1)(n-2)(n-3)3^{n-4}&n=4,\\
n3^{n-1}+\frac{1}{2}n(n-1)3^{n-2}+\frac{2}{3}n(n-1)(n-2)3^{n-3}+\frac{5}{24}n(n-1)(n-2)(n-3)3^{n-4}+\frac{2}{15}n(n-1)(n-2)(n-3)(n-4)3^{n-5}&n\geq 5.
\end{cases}}$