Problem with $0! = 1$

Question

Okay, we know that

$$4!=4 \cdot 3\cdot 2 \cdot 1=24$$ $$5!=5\cdot4\cdot3\cdot2\cdot1=120$$ $$6!=6\cdot5\cdot4\cdot3\cdot2\cdot1=720$$

This is can be written as

$$n!=n\cdot(n-1)!$$

We obtain

$$6! = \frac {7!}{7} = 720$$ $$5! = \frac {6!}{6} = 120$$ $$4! = \frac {5!}{5} = 24$$

If we take $0! = 1$

$$0! = \frac {1!}{1} = \frac {1}{0}$$

Using other way

$$0!=0\cdot0 = 0$$

Why is $0!$ equal to $1$?

And literally 30 other questions... https://math.stackexchange.com/questions/linked/20969 — Asaf Karagila, Oct 09 '17 at 15:07
After spotting your mistake (see Michael), you will conclude $0!=1$. — , Oct 09 '17 at 15:11
Where you have $\dfrac{7!}{7!},$ you need $\dfrac{7!} 7.$ And $\dfrac{1!} 1$ is not $\dfrac 1 0, $ but $\dfrac 1 1,$ so that solves your problem. — Michael Hardy, Oct 09 '17 at 15:11

score 0 · Accepted Answer · answered Oct 09 '17 at 15:06

0

It so that the equality $n!=n\times(n-1)!$ holds for every natural number.

answered Oct 09 '17 at 15:06

Natural numbers do or don't include $0$ depending on the author. – user236182 Oct 09 '17 at 15:35
For me,they don't, but that doesn't invalidate what I wrote. – José Carlos Santos Oct 09 '17 at 15:41

score 0 · Answer 2 · answered Oct 09 '17 at 15:08

0

The core idea of factorial is arranging n numbers so by definition 0 can be arranged in one set only as empty set can be arranged in one way so 0!=1

answered Oct 09 '17 at 15:08

sidhartha

2 Answers2