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Why does $$0!=1$$ ?

 May 10, 2015

Best Answer 

 #1
avatar+870 
+8

It's because $$(n-1)!=\frac{n!}{n}$$

That means :

$$\\6!=1\times 2\times 3\times 4\times 5\times 6=720
\\(6-1)!=5!=\frac{6!}{6}=120
\\(5-1)!=4!=\frac{5!}{5}=24
\\(4-1)!=3!=\frac{4!}{4}=6
\\(3-1)!=2!=\frac{3!}{3}=2
\\(2-1)!=1!=\frac{2!}{2}=1
\\\textcolor[rgb]{0,0,1}{(1-1)!=0!=\frac{1!}{1}=1}$$

 May 10, 2015
 #1
avatar+870 
+8
Best Answer

It's because $$(n-1)!=\frac{n!}{n}$$

That means :

$$\\6!=1\times 2\times 3\times 4\times 5\times 6=720
\\(6-1)!=5!=\frac{6!}{6}=120
\\(5-1)!=4!=\frac{5!}{5}=24
\\(4-1)!=3!=\frac{4!}{4}=6
\\(3-1)!=2!=\frac{3!}{3}=2
\\(2-1)!=1!=\frac{2!}{2}=1
\\\textcolor[rgb]{0,0,1}{(1-1)!=0!=\frac{1!}{1}=1}$$

EinsteinJr May 10, 2015

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