Contoh Soal 2 Kaidah Pencacahan

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

$\begin{array}{ll}7.& \text{Banyaknya cara menyusun huruf-huruf dari}\\ & \text{kata "SEMARANG" adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 1680}& & & \text{d}.& {\displaystyle 20320}\\ \\ \text{b}.& {\displaystyle 6720}& \text{c}.& {\displaystyle 20160}& \text{e}.& {\displaystyle 40320}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Penyelesaian di atas dapat diselesaikan}\\ & \text{baik dengan permutasi maupun kombinasi}\\ & \text{Susunan huruf berbeda yang diambil dari}\\ & \text{kata "SEMARANG" adalah}:\\ & \{\begin{array}{ll}\text{S}& =1\\ \text{E}& =1\\ \text{M}& =1\\ \text{A}& =2\\ \text{R}& =1\\ \text{N}& =1\\ \text{G}& =1\end{array}\\ & \text{Jumlah huruf ada 8 buah}\\ & \text{Dengan cara permutasi}\\ & \begin{array}{rl}P(n;n_1,n_2,n_2,...,n_r)& ={\displaystyle \frac{n!}{n_1!.n_2!.n_3!...n_r!}}\\ P(8;1,1,1,2,1,1,1)& ={\displaystyle \frac{8!}{1!.1!.1!.2!.1!.1!.1!}}\\ & ={\displaystyle \frac{40.320}{2}}\\ & =20.160\end{array}\\ & \text{Dengan cara kombinasi}\\ & \begin{array}{rl}C(n;...)& ={\displaystyle \frac{n!}{n_1!.n_2!.n_3!...n_r!}}\\ C(8;...)& ={\displaystyle (\genfrac{}{}{0ex}{}{8}{1}).(\genfrac{}{}{0ex}{}{7}{1}).(\genfrac{}{}{0ex}{}{6}{1}).(\genfrac{}{}{0ex}{}{5}{2}).(\genfrac{}{}{0ex}{}{3}{1}).(\genfrac{}{}{0ex}{}{2}{1})}\\ & ={\displaystyle 8.7.6.{\displaystyle \frac{5.4}{2}.3.2}}\\ & ={\displaystyle \frac{40.320}{2}}\\ & =20.160\end{array}\end{array}\end{array}$

$\begin{array}{ll}8.& \text{Jumlah susunan dari sebelas huruf}\\ & \mathbf{\text{MISSISSIPPI}}\\ & \text{Banyak susunan berbeda dari semua}\\ & \text{huruf di atas jika keempat huruf}\mathbf{\text{I}}\\ & \text{selalu tampil berdampingan}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle {\displaystyle \frac{9!}{2!4!}}}& & & \text{d}.& {\displaystyle \frac{6!}{2!4!}}\\ \\ \text{b}.& {\displaystyle \frac{8!}{2!4!}}& \text{c}.& {\displaystyle \frac{7!}{2!4!}}& \text{e}.& {\displaystyle \frac{5!}{2!4!}}\end{array}\\ \\ & \text{National University of Singapore}\\ & \text{Sample Test Entrance Examination}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Pandang semua huruf}I\text{dianggap 1}\\ & \text{maka perhitungannnya}\\ & P(8;1,1,4,2)={\displaystyle \frac{8!}{2!4!}}\end{array}\end{array}$

$\begin{array}{ll}9.& \text{Nilai dari}P(4,2)\times P(5,3)=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 12}& & & \text{d}.& {\displaystyle 480}\\ \\ \text{b}.& {\displaystyle 48}& \text{c}.& {\displaystyle 60}& \text{e}.& {\displaystyle 720}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& P(4,2)\times P(5,3)\\ & ={\displaystyle \frac{4!}{(4-2)!}\times \frac{5!}{(5-3)!}}\\ & ={\displaystyle \frac{4!}{2!}\times \frac{5!}{2!}}\\ & ={\displaystyle \frac{4.3.\text{⧸}2!}{\text{⧸}2!}\times \frac{5.4.3.\text{⧸}2!}{\text{⧸}2!}}\\ & =720\end{array}\end{array}$

$\begin{array}{ll}10.& \text{Nilai}n\text{jika}P(n+1,3)=P(n,4)\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 3}& & & \text{d}.& {\displaystyle 6}\\ \\ \text{b}.& {\displaystyle 4}& \text{c}.& {\displaystyle 5}& \text{e}.& {\displaystyle 7}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}P(n+1,3)& =P(n,4)\\ {\displaystyle \frac{(n+1)!}{((n+1)-3)!}}& ={\displaystyle \frac{n!}{(n-4)!}}\\ {\displaystyle \frac{(n+1)!}{(n-2)!}}& =\frac{n!}{(n-4)!}\\ {\displaystyle \frac{(n+1).\text{⧸}n!}{(n-2).(n-3).\text{⧸}(n-4)!}}& ={\displaystyle \frac{\text{⧸}n!}{\text{⧸}(n-4)!}}\\ {\displaystyle \frac{n+1}{n^2-5n+6}}& =1\\ n^2-5n+6& =n+1\\ n^2-6n+5& =0\\ (n-1)(n-5)& =0\\ n=1\text{atau}n=5& \end{array}\end{array}$