Faktorial-Lanjutan Kaidah Pencacahan (Matematika Wajib Kls XII K13 Revisi)

C. Faktorial

Faktorial di sini adalah hasil perkalian semua bilangan asli.

Notasi faktorial ini adalah n! di baca n faktorial.

Untuk setiap bilangan asli n, maka

Didefinisikan pula bahwa \color{blue}0!=1!=1.

Sebagaimana contoh berikut:

\begin{array}{ll}\\ 1.&\textrm{Tentukanlah nilai}\\ &\begin{array}{lll}\\ \textrm{a}.\quad 4!&\textrm{e}.\quad \displaystyle \frac{6!}{4!}&\textrm{i}.\quad \displaystyle \frac{2!}{0!}+\frac{3!}{1!}+\frac{4!}{2!}\\ \textrm{b}.\quad 6!&\textrm{f}.\quad \displaystyle \frac{10!}{6!}&\textrm{j}.\quad \displaystyle \frac{2!}{0!}\times \frac{3!}{1!}+\frac{4!}{2!}\\ \textrm{c}.\quad 0!+1!+2!+3!&\textrm{g}.\quad \displaystyle \frac{7!}{3!\times 4!}&\textrm{k}.\quad \displaystyle \frac{3\times 4!}{3!(5!-5!)}\\ \textrm{d}.\quad (2!)!+(3!)!&\textrm{h}.\quad \displaystyle \frac{13!}{12!+12!}&\textrm{l}.\quad \displaystyle \frac{3!+5!+7!}{4!+6!}\end{array}\\\\ &\textrm{Jawab}:\\\\ &\begin{array}{l}\\ \textrm{a}.\quad 4!=4.3.2.1=24\\ \textrm{b}.\quad 6!=6.5.4.3.2.1=720\\ \begin{aligned}\textrm{c}.\quad 0!+1!+2!+3!&=1+1+2+6\\ &=10 \end{aligned}\\ \begin{aligned}\textrm{d}.\quad (2!)!+(3!)!&=2!+6!\\ &=2+720\\ &=722 \end{aligned}\\ \textrm{e}.\quad \displaystyle \frac{6!}{4!}=\frac{720}{24}=30\quad \textrm{atau}\quad \displaystyle \frac{6!}{4!}=\displaystyle \frac{6.5.\not{4}.\not{3}.\not{2}.\not{1}}{\not{4}.\not{3}.\not{2}.\not{1}}=6.5=30\\ \textrm{f}.\quad \displaystyle \frac{10!}{6!}=\frac{10.9.8.7.6.5.4.3.2.1}{6.5.4.3.2.1}=.... (\textrm{silahkan diselesaikan sendiri})\\ \textrm{g}.\quad \displaystyle \frac{7!}{3!\times 4!}=\frac{7.6.5.4.3.2.1}{(3.2.1)\times (4.3.2.1)}=.... (\textrm{silahkan juga diselesaikan sendiri})\\ \vdots \\ (\textrm{silahkan untuk dicoba sendiri sebagai latihan}) \end{array} \end{array}
\begin{array}{ll}\\ 2.&\textrm{Sederhanakanlah}\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{n!}{(n-1)!}&\textrm{e}.\quad \displaystyle \frac{1}{n!}+\frac{n}{(n+1)!}-\frac{1}{(n-1)!}\\ \textrm{b}.\quad \displaystyle \frac{(n+2)!}{(n+1)!}&\textrm{f}.\quad \displaystyle \frac{(4n)!}{(4n+1)!}+\frac{(4n)!}{(4n-1)!}\\ \textrm{c}.\quad \displaystyle \frac{(2n)!}{(2n+1)!}&\textrm{g}.\quad \displaystyle \frac{1}{n}-\frac{n!}{(n-1).(n-2)!}\\ \textrm{d}.\quad \displaystyle \frac{(n+2)!}{(n^{2}+3n+2)}&\end{array}\\\\ &\textrm{Jawab}:\\\\ &\begin{array}{l}\\ \textrm{a}.\quad \displaystyle \frac{n!}{(n-1)!}=\frac{n.(n-1)!}{(n-1)!}=n\\ \textrm{b}.\quad \displaystyle \frac{(n+2)!}{(n+1)!}=\frac{(n+2).(n+1)!}{(n+1)!}=n+2\\ \textrm{c}.\quad \displaystyle \frac{(2n)!}{(2n+1)!}=\frac{(2n)!}{(2n+1).(2n)!}=\frac{1}{2n+1}\\ \textrm{d}.\quad \displaystyle \frac{(n+2)!}{n^{2}+3n+2}=\frac{(n+2)!}{(n+2).(n+1)}=\frac{(n+2).(n+1).n!}{(n+2).(n+1)}=n!\\ \vdots \\ (\textrm{silahkan dikerjakan sebagai latihan}) \end{array} \end{array}

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