$\textcolor{ForestGreen}{\textbf{1. $^nC_0 = {}^nC_n = 1$}}$  

Penurunan:  

$$
^nC_0 = \frac{n!}{0!\,(n-0)!}
       = \frac{n!}{1\cdot n!} = 1,
\quad
^nC_n = \frac{n!}{n!\,0!}
       = \frac{n!}{n!\cdot1} = 1.
$$

Contoh:  

$$
^5C_0 = 1,\quad ^5C_5 = 1.
$$


$\textcolor{ForestGreen}{\textbf{2. $^nC_r = {}^nC_{n-r}$}}$  

Penurunan:  

$$
^nC_r
= \frac{n!}{r!\,(n-r)!}
= \frac{n!}{(n-r)!\,r!}
= {}^nC_{n-r}.
$$

Contoh:  

$$
^7C_2 = \frac{7!}{2!\,5!} = 21,
\quad
^7C_5 = \frac{7!}{5!\,2!} = 21.
$$


$\textcolor{ForestGreen}{\textbf{3. Jika $^nC_r = {}^nC_k$ maka $r = k$ atau $n-r = k$}}$  

Penurunan:  

$$
r = k \quad\text{atau}\quad n-r = k.
$$

Contoh:  

$$
^6C_2 = 15,\quad ^6C_4 = 15
\;\Longrightarrow\;2+4=6.
$$


$\textcolor{ForestGreen}{\textbf{4. $^nC_r + {}^nC_{r-1} = {}^{n+1}C_r$}}$  

Penurunan:  

$$
\frac{n!}{r!\,(n-r)!}
+
\frac{n!}{(r-1)!\,(n-r+1)!}
=
\frac{(n+1)!}{r!\,(n+1-r)!}.
$$

Contoh:  

$$
^5C_2 + {}^5C_1 = 10 + 5 = 15 = {}^6C_2.
$$


$\textcolor{ForestGreen}{\textbf{5. $r\cdot {}^nC_r = n\cdot {}^{n-1}C_{r-1}$}}$  

Penurunan:  

$$
r\;\frac{n!}{r!\,(n-r)!}
=
n\;\frac{(n-1)!}{(r-1)!\,(n-r)!}
= n\;{}^{n-1}C_{r-1}.
$$

Contoh:  

$$
2\cdot {}^5C_2 = 20,
\quad
5\cdot {}^4C_1 = 20.
$$


$\textcolor{ForestGreen}{\textbf{6. $\displaystyle \frac{1}{r+1}\,{}^nC_r = \frac{1}{n+1}\,{}^{n+1}C_{r+1}$}}$  

Penurunan:  

$$
\frac{1}{r+1}\,\frac{n!}{r!\,(n-r)!}
=
\frac{1}{n+1}\,\frac{(n+1)!}{(r+1)!\,(n-r)!}.
$$

Contoh:  

$$
\tfrac{1}{3}\,^5C_2 = \tfrac{10}{3},
\quad
\tfrac{1}{6}\,^6C_3 = \tfrac{10}{3}.
$$


$\textcolor{ForestGreen}{\textbf{7. $\displaystyle \frac{{}^nC_r}{{}^nC_{r-1}} = \frac{n - r + 1}{r}$}}$  

Penurunan:  

$$
\frac{{}^nC_r}{{}^nC_{r-1}}
=
\frac{n-r+1}{r}.
$$

Contoh:  

$$
\frac{{}^7C_3}{{}^7C_2} = \frac{35}{21} = \frac{5}{3}.
$$


$\textcolor{ForestGreen}{\textbf{8. Nilai maksimum $^nC_r$}}$  

Penurunan: bandingkan rasio  

$$
\frac{{}^nC_r}{{}^nC_{r-1}} = \frac{n-r+1}{r},
$$  

maksimum terjadi saat rasio turun di bawah 1.

Contoh:  

$n=6$: baris ke-6 Pascal adalah 1,6,15,20,15,6,1 → maksimum 20 di $r=3$.  

$n=7$: baris ke-7 adalah 1,7,21,35,35,21,7,1 → maksimum 35 di $r=3$ dan $r=4$.