Why is the number of arrangements of $n$ objects equal to $n!$?

Question

I hope this question does not seem vague, but whats the logic/reason behind that e.g. One can arrange 5 colours in $5!$ different ways if all of the colours are different?

Answer 1

Suppose we put 5 different colors in a queue. The first position has 5 choices. Once you have picked the first color, you have 4 colors left. So there are 4 choices for the second position, so on and so forth. Hence, you have $5\cdot4\cdot3\cdot2\cdot1=5!$ different arrangements.

Answer 2

The idea is to look at the recursion: The number of arrangements that start which each of the $5$ colors is the same.

Therefore the answer is $5$ multiplied by the number of ways to arrange $4$ colors. And the number of ways to order $4$ colors is $4$ multiplied by the number of ways to order $3$ colors $\dots$

So if $f(n)$ is the number of ways to arrange $n$ colors we have $f(n+1)=(n+1)f(n)$. And since the number of ways to order $1$ color is $1$ we have our answer.

Answer 3

Suppose I have (to make things concrete) $5$ slips of paper, each of a different color; I'm to put these in some order. I can take any one of the $5$ and put it first. For each of those $5$, I have a choice of $4$ to put second; that's a total of

$(4 + 4 + 4 + 4 + 4)$ or $5\cdot4 = 20$

ways to arrange the first two slips.

For each of these $20$ different arrangements, I can put $3$ different slips third:

$(3 + 3 + 3 + ... + 3)$ (twenty times)
$(20\cdot3) = 60$
$(5\cdot4\cdot3) = 60$

ways to arrange the first three slips.

In a similar way, I have $2$ slips to put down fourth, and

$(5\cdot4\cdot3\cdot2) = 120$

ways to arrange these first four slips.

Then there's only one slip left, which I must put down last. I have no choice (or if you like, a choice of $1$ possibility) in this matter. That gives me

$(5\cdot4\cdot3\cdot2\cdot1) = 120 = 5!$

possible ways to arrange the $5$ slips.