Evaluating \(\displaystyle{ n\choose{j}}\)

1. \(\displaystyle{{3}\choose{1}} =\dfrac{3!}{1! ( 3-1) !}=\dfrac{3!}{1!2!}=\dfrac{3\cdot 2\cdot 1}{1 ( 2\cdot 1) }=\dfrac{6}{2}=3\)
2. \(\displaystyle{{4}\choose{2}} =\dfrac{4!}{2! ( 4-2) !}=\dfrac{4!}{2!2!}=\dfrac{4\cdot 3\cdot 2\cdot 1}{ ( 2\cdot 1) ( 2\cdot 1) }=\dfrac{24}{4}=6\)
3. \(\displaystyle{{8}\choose{7}} =\dfrac{8!}{7! ( 8-7) !}=\dfrac{8!}{7!1!} \underset{\underset{\color{#0066A7}{\hbox{\(8!=8\cdot 7!\)}}}{\color{#0066A7}{\displaystyle\uparrow}}}{=} \dfrac{8\cdot 7!}{7!\cdot 1!}=\dfrac{8}{1}=8\)
4. \(\displaystyle{{n}\choose{0}} =\dfrac{n!}{0! ( n-0) !}=\dfrac{n!}{0!n!}=\dfrac{n!}{1\cdot n!}=1\)
5. \(\displaystyle{{n}\choose{n}} =\dfrac{n!}{n! ( n-n) !}=\dfrac{n!}{n!0!}=\dfrac{n!}{n!\cdot 1}=1 \)