Contoh Soal 1 Kaidah Pencacahan

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

$\begin{array}{ll}\ 2.&\textrm{Permutasi 4 unsur dari 11 unsur}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle 7980&&&\textrm{d}.&\displaystyle 7290\\\\ \textrm{b}.&\color{red}\displaystyle 7920&\textrm{c}.&\displaystyle 7820&\textrm{e}.&\displaystyle 7280 \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}P(n,r)&=\displaystyle \frac{n!}{(n-r)!}\\ P(11,4)&=\displaystyle \frac{11!}{(11-4)!}\\ &=\displaystyle \frac{11!}{7!}=\frac{11\times 10\times 9\times 8\times \not{7!}}{\not{7!}}\\ &=\color{red}7920 \end{aligned} \end{array}$

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

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

$\begin{array}{ll}\ 5.&\textrm{Banyaknya cara milih 4 orang dari 10 orang }\\ &\textrm{anggota jika salah seorang di antaranya}\\ &\textrm{selalu terpilih adalah}.... \\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle 72&&&\textrm{d}.&\displaystyle 504\\\\ \color{red}\textrm{b}.&\color{red}\displaystyle 84&\textrm{c}.&\displaystyle 252&\textrm{e}.&\displaystyle 3024 \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Cara memilih}=\textrm{Kombinasi}=C(10-1,4-1)\\ & \textrm{karena 1 orang di antaranya selalu ada/terpilih}\\ &=C(9,3)\\ &=\binom{9}{3}\\ &=\displaystyle \frac{9!}{3!\times (9-3)!}\\ &=\displaystyle \frac{9\times 8\times 7\times \not{6!}}{3\times 2\times \times \not{6!}}\\&=\displaystyle \frac{9.8.7}{3.2}\\ &=\color{red}84 \end{aligned} \end{array}$