Contoh Soal 2 Kaidah Pencacahan

 $\begin{array}{ll}\ 6.&\textrm{Banyaknya cara milih 4 orang dari 10 orang }\\ &\textrm{anggota jika salah seorang di antaranya}\\ &\textrm{selalu terpilih adalah}.... \\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle 72&&&\textrm{d}.&\displaystyle 504\\\\ \color{red}\textrm{b}.&\color{red}\displaystyle 84&\textrm{c}.&\displaystyle 252&\textrm{e}.&\displaystyle 3024 \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Cara memilih}=\textrm{Kombinasi}=C(10-1,4-1)\\ & \textrm{karena 1 orang di antaranya selalu ada/terpilih}\\ &=C(9,3)\\ &=\binom{9}{3}\\ &=\displaystyle \frac{9!}{3!\times (9-3)!}\\ &=\displaystyle \frac{9\times 8\times 7\times \not{6!}}{3\times 2\times \times \not{6!}}\\&=\displaystyle \frac{9.8.7}{3.2}\\ &=\color{red}84 \end{aligned} \end{array}$

$\begin{array}{ll}\ 7.&\textrm{Banyaknya cara menyusun huruf-huruf dari}\\ &\textrm{kata "SEMARANG" adalah}\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle 1680&&&\textrm{d}.&\displaystyle 20320\\\\ \textrm{b}.&\displaystyle 6720&\textrm{c}.&\color{red}\displaystyle 20160&\textrm{e}.&\displaystyle 40320 \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Penyelesaian di atas dapat diselesaikan}\\ &\textrm{baik dengan permutasi maupun kombinasi}\\ &\textrm{Susunan huruf berbeda yang diambil dari}\\ &\textrm{kata "SEMARANG" adalah}:\\ &\begin{cases} \textrm{S} &=1 \\ \textrm{E} &=1 \\ \textrm{M} &=1 \\ \textrm{A} &=2 \\ \textrm{R} &=1 \\ \textrm{N} &=1 \\ \textrm{G} &=1 \end{cases}\\ &\textrm{Jumlah huruf ada 8 buah}\\ &\color{purple}\textrm{Dengan cara permutasi}\\ &\begin{aligned}P(n;n_{1},n_{2},n_{2},...,n_{r})&=\displaystyle \frac{n!}{n_{1}!.n_{2}!.n_{3}!...n_{r}!}\\ P(8;1,1,1,2,1,1,1)&=\displaystyle \frac{8!}{1!.1!.1!.2!.1!.1!.1!}\\ &=\displaystyle \frac{40.320}{2}\\ &=\color{red}20.160 \end{aligned}\\ &\color{purple}\textrm{Dengan cara kombinasi}\\ &\begin{aligned}C(n;...)&=\displaystyle \frac{n!}{n_{1}!.n_{2}!.n_{3}!...n_{r}!}\\ C(8;...)&=\displaystyle \binom{8}{1}.\binom{7}{1}.\binom{6}{1}.\binom{5}{2}.\binom{3}{1}.\binom{2}{1}\\ &=\displaystyle 8.7.6.\displaystyle \frac{5.4}{2}.3.2\\ &=\displaystyle \frac{40.320}{2}\\ &=\color{red}20.160 \end{aligned} \end{aligned} \end{array}$

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

$\begin{array}{ll}\ 9.&\textrm{Nilai dari}\: \: P(4,2)\times P(5,3)=\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle 12&&&\textrm{d}.&\displaystyle 480\\\\ \textrm{b}.&\displaystyle 48&\textrm{c}.&\displaystyle 60&\textrm{e}.&\color{red}\displaystyle 720 \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&P(4,2)\times P(5,3)\\ &=\displaystyle \frac{4!}{(4-2)!}\times \frac{5!}{(5-3)!}\\ &=\displaystyle \frac{4!}{2!}\times \frac{5!}{2!}\\ &=\displaystyle \frac{4.3.\not{2!}}{\not{2!}}\times \frac{5.4.3.\not{2!}}{\not{2!}}\\ &=\color{red}720 \end{aligned} \end{array}$

$\begin{array}{ll}\ 10.&\textrm{Nilai}\: \: n\: \: \textrm{jika}\: \: P(n+1,3)=P(n,4)\\ &\textrm{adalah}\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle 3&&&\textrm{d}.&\displaystyle 6\\\\ \textrm{b}.&\displaystyle 4&\textrm{c}.&\color{red}\displaystyle 5&\textrm{e}.&\displaystyle 7 \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}P(n+1,3)&=P(n,4)\\ \displaystyle \frac{(n+1)!}{((n+1)-3)!}&=\displaystyle \frac{n!}{(n-4)!}\\ \displaystyle \frac{(n+1)!}{(n-2)!}&=\frac{n!}{(n-4)!}\\ \displaystyle \frac{(n+1).\not{n!}}{(n-2).(n-3).\not{(n-4)!}}&=\displaystyle \frac{\not{n!}}{\not{(n-4)!}}\\ \displaystyle \frac{n+1}{n^{2}-5n+6}&=1\\ n^{2}-5n+6&=n+1\\ n^{2}-6n+5&=0\\ (n-1)(n-5)&=0\\ n=1\: \: \textrm{atau}\: \: n=\color{red}5& \end{aligned} \end{array}$