+26400 factorial definition:
$$\boxed{ n! = n \cdot (n-1)! }$$
Example:
$$\\5! = 5 \cdot 4! \\
4! = 4 \cdot 3! \\
3! = 3 \cdot 2! \\
2! = 2 \cdot 1! \\
1! = 1 \cdot 0! \\
0! = 1 ~ \text{(convention)}\\
\boxed{\,5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot \textcolor[rgb]{150,0,0}{ 1 } \,}$$
+118723 There are several answers to this.
I could say it is by convention.
or I could say because it works.
Think about this. 3! is the number of ways you can line 3 people up in a queue
2! is the number of ways that you can line up 2 people A then b or B then A
1! is the number of ways you can line up 1 person.
How many ways can you line up 0 people. Well 1 way. There is only one way that you can put nobody in a queque
0!=1
I am sure that there are many more explanations but I guess the most important one is that it makes probablility questions work.
Lame perhaps but true :)
+26400 factorial definition:
$$\boxed{ n! = n \cdot (n-1)! }$$
Example:
$$\\5! = 5 \cdot 4! \\
4! = 4 \cdot 3! \\
3! = 3 \cdot 2! \\
2! = 2 \cdot 1! \\
1! = 1 \cdot 0! \\
0! = 1 ~ \text{(convention)}\\
\boxed{\,5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot \textcolor[rgb]{150,0,0}{ 1 } \,}$$
