Contoh Soal 1 Kaidah Pencacahan

$\begin{array}{ll}1.& \text{Nilai dari}{\displaystyle \frac{1}{14!}-\frac{10}{15!}+\frac{4}{16!}}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{114}{16!}}& & & \text{d}.& {\displaystyle \frac{9}{16!}}\\ \\ \text{b}.& {\displaystyle \frac{108}{16!}}& \text{c}.& {\displaystyle \frac{84}{16!}}& \text{e}.& {\displaystyle \frac{4}{16!}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}{\displaystyle \frac{1}{14!}}& -{\displaystyle \frac{10}{15!}+\frac{4}{16!}}\\ & ={\displaystyle \frac{15\times 16}{14!\times 15\times 16}-\frac{10\times 16}{15!\times 16}+\frac{4}{16!}}\\ & ={\displaystyle \frac{240}{16!}-\frac{160}{16!}+\frac{4}{16!}}\\ & ={\displaystyle \frac{84}{16!}}\end{array}\end{array}$

$\begin{array}{ll}2.& \text{Permutasi 4 unsur dari 11 unsur}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 7980}& & & \text{d}.& {\displaystyle 7290}\\ \\ \text{b}.& {\displaystyle 7920}& \text{c}.& {\displaystyle 7820}& \text{e}.& {\displaystyle 7280}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}P(n,r)& ={\displaystyle \frac{n!}{(n-r)!}}\\ P(11,4)& ={\displaystyle \frac{11!}{(11-4)!}}\\ & ={\displaystyle \frac{11!}{7!}=\frac{11\times 10\times 9\times 8\times \text{⧸}7!}{\text{⧸}7!}}\\ & =7920\end{array}\end{array}$

$\begin{array}{ll}3.& \text{Empat siswa dan dua siswi akan duduk}\\ & \text{berdampingan. Apabila siswi selalu duduk}\\ & \text{paling pinggir, banyak cara mereka duduk}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 24}& & & \text{d}.& {\displaystyle 64}\\ \text{b}.& {\displaystyle 48}& \text{c}.& {\displaystyle 56}& \text{e}.& {\displaystyle 72}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Total ada 6 anak; 4 siswa, 2 siswi}\\ & \text{Karena ini posisi orang, maka dan semuanya}\\ & \text{tidak identik, maka dapat diurutkan}\\ & \text{Sehingga rumus yang dipergunakan adalah}\\ & \text{permutasi, yaitu}:\\ & \text{Perhatikan posisi mereka}\\ & \mathbf{\text{Posisi pertama}}\\ & \begin{array}{|cccccc|}\hline (1)& (2)& (3)& (4)& (5)& (6)\\ \text{A}& ◻& ◻& ◻& ◻& \text{B}\\ \hline\end{array}\\ & =P(1,1)\times P(4,4)\times P(1,1)=24\\ & \mathbf{\text{Posisi kedua}}\\ & \begin{array}{|cccccc|}\hline (1)& (2)& (3)& (4)& (5)& (6)\\ \text{B}& ◻& ◻& ◻& ◻& \text{A}\\ \hline\end{array}\\ & =P(1,1)\times P(4,4)\times P(1,1)=24\\ & \text{Total}=24+24=48\end{array}\end{array}$

$\begin{array}{ll}4.& \text{Jika}P(7,r)=210,\text{maka nilai}r\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 2}& & & \text{d}.& {\displaystyle 5}\\ \\ \text{b}.& {\displaystyle 3}& \text{c}.& {\displaystyle 4}& \text{e}.& {\displaystyle 6}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}P(7,r)& ={\displaystyle \frac{7!}{(7-r)!}}\\ 210& ={\displaystyle \frac{7!}{(7-r!)}}\\ (7-r)!& ={\displaystyle \frac{7!}{210}=\frac{7\times 6\times 5\times 4\times 3\times 2\times 1}{7\times 5\times 3\times 2\times 1}}\\ (7-r)!& =6.4=24\\ (7-r)!& =4!\\ 7-r& =4\\ r& =7-4\\ r& =3\end{array}\end{array}$

$\begin{array}{ll}5.& \text{Banyaknya cara milih 4 orang dari 10 orang }\\ & \text{anggota jika salah seorang di antaranya}\\ & \text{selalu terpilih adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 72}& & & \text{d}.& {\displaystyle 504}\\ \\ \text{b}.& {\displaystyle 84}& \text{c}.& {\displaystyle 252}& \text{e}.& {\displaystyle 3024}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Cara memilih}=\text{Kombinasi}=C(10-1,4-1)\\ & \text{karena 1 orang di antaranya selalu ada/terpilih}\\ & =C(9,3)\\ & =(\genfrac{}{}{0ex}{}{9}{3})\\ & ={\displaystyle \frac{9!}{3!\times (9-3)!}}\\ & ={\displaystyle \frac{9\times 8\times 7\times \text{⧸}6!}{3\times 2\times \times \text{⧸}6!}}\\ & ={\displaystyle \frac{9.8.7}{3.2}}\\ & =84\end{array}\end{array}$