Contoh Soal 3 Kaidah Pencacahan

 $\begin{array}{ll}\ 11.&\textrm{Berikut ini nilainya tidak sama dengan}\\ &C(7,5)\: \: \textrm{adalah}\: ....\\\\ &(i)\: \: \displaystyle \frac{7!}{5!(7-5)!}\\\\ &(ii)\: \: C(6,1)\\\\ &(iii)\: \: \displaystyle \frac{P(7,5)}{5!}\\\\ &(iv)\: \: \begin{pmatrix} 6\\ 1 \end{pmatrix}\\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle (i),(ii),\& (iii)&&&\textrm{d}.&\displaystyle \textrm{hanya}\: (i)\\\\ \textrm{b}.&\displaystyle (i)\& (iii)&\textrm{c}.&\color{red}(ii)\&(iv)&\textrm{e}.&\displaystyle \textrm{hanya}\: (iv) \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &C(7,5)=\displaystyle \frac{P(7,5)}{5!}=\frac{7!}{5!(7-5)!} \end{array}$

$\begin{array}{ll}\ 12.&\textrm{Nilai}\: \: n\: \: \textrm{yang memenuhi persamaan}\\ &\begin{pmatrix} 100\\ 45 \end{pmatrix}=\begin{pmatrix} 100\\ 5n \end{pmatrix}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle 15&&&\textrm{d}.&\displaystyle 12\\\\ \textrm{b}.&\displaystyle 14&\textrm{c}.&\displaystyle 13&\textrm{e}.&\color{red}\displaystyle 11 \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}\\ &\begin{pmatrix} 100\\ 45 \end{pmatrix}=\begin{pmatrix} 100\\ 5n \end{pmatrix},\: \: \textrm{maka}\\ &45+5n=100\\ &5n=100-45=55\\ &\: \: n=\displaystyle \frac{55}{5}=\color{red}11 \end{aligned} \end{array}$

$\begin{array}{ll}\ 13.&\textrm{Koefisien suku ke-4 dari}\: \: (2x-3)^{4}\\ &\begin{array}{llllll}\\ \textrm{a}.&\color{red}\displaystyle -216&&&\textrm{d}.&\displaystyle 81\\\\ \textrm{b}.&\displaystyle -96&\textrm{c}.&\displaystyle 16&\textrm{e}.&\displaystyle 216 \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}\\ &(2x-3)^{4}=\displaystyle \sum_{i=0}^{4}\begin{pmatrix} 4\\ i \end{pmatrix}(2x)^{4-i}(-3)^{i}\\ &\textrm{Suku ke-4-nya adalah}:\: \: r=4.\\ &\textrm{Suku ke-r}=\color{red}\begin{pmatrix} n\\ r-1 \end{pmatrix}a^{n-r+1}b^{r-1}\\ &\textrm{Sehingga suku ke-4 adalah}:\\ &=\begin{pmatrix} 4\\ 4-1 \end{pmatrix}(2x)^{4-4+1}(-3)^{4-1}\\ &=\begin{pmatrix} 4\\ 3 \end{pmatrix}(2x)^{1}(-3)^{3}\\ &=\displaystyle \frac{4!}{3!\times 1!}2x(-27)\\ &=-4.2.27x\\ &=\color{red}-216 \end{aligned} \end{array}$

$\begin{array}{ll}\ 14.&\textrm{Bentuk sederhana dari}\: \: \displaystyle \sum_{r=1}^{n}r\displaystyle \begin{pmatrix} n\\ r \end{pmatrix}\\ &\textrm{dengan}\: \: \begin{pmatrix} n\\ r \end{pmatrix}=\displaystyle \frac{n!}{r!(n-r)!}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle 2^{n+1}&&&\textrm{d}.&\displaystyle 3^{n}\\\\ \textrm{b}.&\color{red}\displaystyle n2^{n-1}&\textrm{c}.&\displaystyle n2^{n}&\textrm{e}.&\displaystyle 3^{n+1} \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\displaystyle \sum_{r=1}^{n}r\displaystyle \begin{pmatrix} n\\ r \end{pmatrix}&=\displaystyle \sum_{r=1}^{n}r\displaystyle \frac{n!}{r!(n-r)!}\\ &=\displaystyle \sum_{r=1}^{n}r\displaystyle \frac{n(n-1)!}{r(r-1)!(n-r)!}\\ &=n\displaystyle \sum_{r=1}^{n}\displaystyle \frac{(n-1)!}{(r-1)!(n-r)!}\\ &=n\displaystyle \sum_{r=1}^{n}\displaystyle \frac{(n-1)!}{(r-1)!((n-1)-(r-1))!}\\ &=\displaystyle \sum_{r=1}^{n}\begin{pmatrix} n-1\\ r-1 \end{pmatrix}\\ &=\color{red}n.2^{r-1} \end{aligned} \end{array}$

$\begin{array}{ll}\ 15.&\textrm{Banyaknya diagonal segi 6 adalah}\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle 15&&&\textrm{d}.&\color{red}\displaystyle 9\\\\ \textrm{b}.&\displaystyle 14&\textrm{c}.&\displaystyle 10&\textrm{e}.&\displaystyle 6 \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Banyak diagonal segi}-n \: \: \textrm{adalah}:\\ &C(n,2)-n.\: \textrm{Jika seperti soal dengan}\\ &n=6,\: \: \textrm{maka}\\ &C(6,2)=\displaystyle \frac{6!}{2!\times 4!}=\frac{6\times 5\times \not{4!}}{2\times 1\times \not{4!}}=15\\ &\textrm{Sehingga}\\ &C(6,2)-6=15-6=\color{red}9 \end{aligned} \end{array}$