Contoh Soal 3 Kaidah Pencacahan

$\begin{array}{ll}11.& \text{Nilai}n\text{yang memenuhi persamaan}\\ & \left(\begin{array}{c}100\\ 45\end{array}\right)=\left(\begin{array}{c}100\\ 5n\end{array}\right)\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 15}& & & \text{d}.& {\displaystyle 12}\\ \\ \text{b}.& {\displaystyle 14}& \text{c}.& {\displaystyle 13}& \text{e}.& {\displaystyle 11}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & \left(\begin{array}{c}100\\ 45\end{array}\right)=\left(\begin{array}{c}100\\ 5n\end{array}\right),\text{maka}\\ & 45+5n=100\\ & 5n=100-45=55\\ & n={\displaystyle \frac{55}{5}=11}\end{array}\end{array}$

$\begin{array}{ll}12.& \text{Koefisien suku ke-4 dari}(2x-3{)}^4\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -216}& & & \text{d}.& {\displaystyle 81}\\ \\ \text{b}.& {\displaystyle -96}& \text{c}.& {\displaystyle 16}& \text{e}.& {\displaystyle 216}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & (2x-3{)}^4={\displaystyle \sum _{i=0}^4\left(\begin{array}{c}4\\ i\end{array}\right)(2x{)}^{4-i}(-3{)}^i}\\ & \text{Suku ke-4-nya adalah}:r=4.\\ & \text{Suku ke-r}=\left(\begin{array}{c}n\\ r-1\end{array}\right)a^{n-r+1}{b}^{r-1}\\ & \text{Sehingga suku ke-4 adalah}:\\ & =\left(\begin{array}{c}4\\ 4-1\end{array}\right)(2x{)}^{4-4+1}(-3{)}^{4-1}\\ & =\left(\begin{array}{c}4\\ 3\end{array}\right)(2x{)}^1(-3{)}^3\\ & ={\displaystyle \frac{4!}{3!\times 1!}2x(-27)}\\ & =-4.2.27x\\ & =-216\end{array}\end{array}$

$\begin{array}{ll}13.& \text{Bentuk sederhana dari}{\displaystyle \sum _{r=1}^{n}r{\displaystyle \left(\begin{array}{c}n\\ r\end{array}\right)}}\\ & \text{dengan}\left(\begin{array}{c}n\\ r\end{array}\right)={\displaystyle \frac{n!}{r!(n-r)!}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 2^{n+1}}& & & \text{d}.& {\displaystyle 3^n}\\ \\ \text{b}.& {\displaystyle n{2}^{n-1}}& \text{c}.& {\displaystyle n{2}^n}& \text{e}.& {\displaystyle 3^{n+1}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}{\displaystyle \sum _{r=1}^{n}r{\displaystyle \left(\begin{array}{c}n\\ r\end{array}\right)}}& ={\displaystyle \sum _{r=1}^{n}r{\displaystyle \frac{n!}{r!(n-r)!}}}\\ & ={\displaystyle \sum _{r=1}^{n}r{\displaystyle \frac{n(n-1)!}{r(r-1)!(n-r)!}}}\\ & =n{\displaystyle \sum _{r=1}^n{\displaystyle \frac{(n-1)!}{(r-1)!(n-r)!}}}\\ & =n{\displaystyle \sum _{r=1}^n{\displaystyle \frac{(n-1)!}{(r-1)!((n-1)-(r-1))!}}}\\ & ={\displaystyle \sum _{r=1}^n\left(\begin{array}{c}n-1\\ r-1\end{array}\right)}\\ & =n{.2}^{r-1}\end{array}\end{array}$

$\begin{array}{ll}14.& \text{Banyaknya diagonal segi 6 adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 15}& & & \text{d}.& {\displaystyle 9}\\ \\ \text{b}.& {\displaystyle 14}& \text{c}.& {\displaystyle 10}& \text{e}.& {\displaystyle 6}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Banyak diagonal segi}-n\text{adalah}:\\ & C(n,2)-n.\text{Jika seperti soal dengan}\\ & n=6,\text{maka}\\ & C(6,2)={\displaystyle \frac{6!}{2!\times 4!}=\frac{6\times 5\times \text{⧸}4!}{2\times 1\times \text{⧸}4!}=15}\\ & \text{Sehingga}\\ & C(6,2)-6=15-6=9\end{array}\end{array}$

$\begin{array}{ll}15.& \text{Diketahui himpunan yang terdiri dari 5}\\ & \text{huruf vokal dan 10 huruf konsonan yang}\\ & \text{semuanya berlainan. Dari himpunan itu}\\ & \text{disusun suatu kata yang terdiri dari 2}\\ & \text{huruf vokal dan 3 konsonan. Banyak kata}\\ & \text{yang dapat disusun sebanyak}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 144.000}& & & \text{d}.& {\displaystyle 72.000}\\ \\ \text{b}.& {\displaystyle 126.000}& \text{c}.& {\displaystyle 96.000}& \text{e}.& {\displaystyle 36.000}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui bahwa ingin menyusun}\\ & \mathbf{\text{5 huruf dengan susunan berbeda}}\\ & \mathit{\text{yang tersusun dari}}\\ & \text{2 dari 5 vokal berbeda disusun, dan}\\ & \text{3 dari 10 konsonan berbeda juga disusun}\\ & \text{maka banyak susunan kata terbentuk}:\\ & \text{Seperti menyusun 5 objek (kombinasi)}\\ & \text{2 benda dari 5 benda, atau 3 }\\ & \text{benda yang terbentuk dari 5}\\ & \text{objek yg tidak identik(permutasi)}\\ & \mathbf{\text{Cara pertama}}\\ & =C((2+3),2)\times P(5,2)\times P(10,5)\\ & ={\displaystyle \frac{5!}{2!\times 3!}\times \frac{5!}{(5-2)!}\times \frac{10!}{(10-3)!}}\\ & ={\displaystyle \frac{5!}{2!\times 3!}\times \frac{5!}{3!}\times \frac{10!}{7!}}\\ & =10\times 60\times 720\\ & =144.000\\ & \mathbf{\text{Cara kedua}}\\ & =C((2+3),3)\times P(5,2)\times P(10,5)\\ & ={\displaystyle \frac{5!}{3!\times 2!}\times \frac{5!}{(5-2)!}\times \frac{10!}{(10-3)!}}\\ & ={\displaystyle \frac{5!}{3!\times 2!}\times \frac{5!}{3!}\times \frac{10!}{7!}}\\ & =10\times 60\times 720\\ & =144.000\end{array}\end{array}$