Below, you'll find five unconventional and different ways to explain why 0! is 1
The first way: is by using the Combinatorial Argument: Combinatorics, in general, looks at the different number of ways to arrange a number (n) of other objects. When n is 0, there are simply no objects to arrange, but there is, one way to arrange nothing, that is, of course, to do nothing! This is the only arrangement possible, so by definition (0! = 1). Easy enough, right?
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Second way is by recursive (rekürsif fonksiyon)
Factorial is recursively defined with its base case at (0!). For any integer (n > 0), (n! = n \times (n-1)!). Applying that definition backward, (1! = 1 \times 0!), and since (1! = 1), it shows us that (0! = 1) and additionally supports our argument.
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Third way is by Algebraic Proof with the use of Permutations: Consider the formula for permutations of (n) items taken (n) at a time, which is given by (n!/(n! \times 0!)). If we take (n = 3), we get (3!/(3! \times 0!) = 1). Solving for (0!), we see that (3! \times 0! = 3!), which means(0! = 1).
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Fourth way is by Gamma Function: The Gamma function (\Gamma(n)) is an extension of the factorial function to complex numbers for Γ(n) = (n-1)! for positive integers. Observe that the Gamma function is continuous and Γ(1) = 0! But since Γ(1) = 1, we must conclude that 0! = 1.
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The final way is: The Empty Product Convention: In mathematics, the result of multiplying no numbers at all is called the multiplicative identity because identity functions leave a number unchanged. This is expressed with the empty product convention.
Since (0!) is precisely the product of no numbers, according to this way as well, it equals 1. These explanations gather mathematical definitions, properties of functions, and conventions to prove that 0! equals 1. It is a simple and a very fundamental concept that ensures consistency across a wide variety of mathematical disciplines.
I hope that clears up the confusion many students get when they learn that 0! equals 1.
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