Most recent change of ZeroFactorial

Edit made on February 26, 2009 by RiderOfGiraffes at 09:28:13

Deleted text in red / Inserted text in green

WW
!! The question ...
What "should" zero factorial be?

There are several explanations.

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COLUMN_START
[[[
!! Direct - number of arrangements
Working directly from the definition of factorial as the
number of arrangements of objects, we declare that the
number of ways of arranging zero objects is one, but some
people are unconvinced by this. "There are no objects",
they say, "so you can't arrange them at all. The answer
should be zero."

Well, here's an elaboration of the "arrangement" explanation.

* Suppose you have 5 boxes and five objects. One person goes into the room and puts one thing in each box. Then someone else goes in, and they have to write down what the arrangement is. How many different answers can you get? 5! = 120.

* Suppose you have 4 boxes and four objects. One person goes into the room and puts one thing in each box. Then someone else goes in, and they have to write down what the arrangement is. How many different answers can you get? 4! = 24.

* ...

* Suppose you have one box and one object. One person goes into the room and puts one thing in each box. Then someone else goes in, and they have to write down what the arrangement is. How many different answers can you get? 1! = 1.

Now for zero boxes? There is only one answer that can be given.
]]] _ [[[
!! Identity approach
If we define EQN:n! to be EQN:n(n-1)(n-2)...3.2.1 then we
can deduce that EQN:\frac{n!}{(n-1)!}=n. Substituting n=1 we get that
EQN:\frac{1!}{0!}=1, and hence 0!=1.
]]] _ [[[
!! Another identity
Since the number of ways of choosing r objects from n is
EQN:\frac{n!}{r!(n-r)!} the number of ways of choosing n objects from
n objects is EQN:\frac{n!}{n!.0!}, so 0! must be 1 to make that work.
]]]

COLUMN_SPLIT
[[[
!! The number pattern explanation
Look at the following sequence:

40320, 5040, 720, 120, 24, ...

Ask "How does this continue? What's the rule?"

Some work will show that going leftwards involves multiplying
by successive integers, so going rightwards should be dividing
by successive integers:
{{{
40320 (divide by 8 gives ...)
5040 (divide by 7 gives ... )
720 (divide by 6 gives ...)
120 (divide by 5 gives ...)
24 (divide by 4 gives ...)
??
}}}
This shows clearly the pattern we want.
]]] _ [[[
!! The "Algorithm" approach.
Here's an algorithm to compute /n/ factorial (written in Python (yes, this is executable code))
{{{
. F = 1
. while n>0:
. . F = F*n
. . n = n-1
. print F
}}}
Running through this a few times shows how it works, why it works, and that it gives the right answer for n>0. What answer does it give for n=0?
]]] _ [[[
!! Another identity
Since the number of ways of choosing r objects from n is
EQN:\frac{n!}{r!(n-r)!} the number of ways of choosing n objects from
n objects is EQN:\frac{n!}{n!.0!}, so 0! must be 1 to make that work.
]]]

COLUMN_END
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!! Conclusion
Yup. 0! = 1.

!! Credits
Some of these have been suggested by contributors on the TES
teacher bulletin board: