The digits 3, 4. 5, and 6 are used to form a four digit code. No repeated digits are allowed. How many codes are possible?

Question

I'm not exactly sure how to solve this specific problem, but I think I would need to come up with 2 or more equations then solve them together... can't come up with equations though. Can I get some help?

Answer 1

No equations are needed. Let's construct the number of codes:

We have $4$ numbers at our disposal. Let's take a number and make it our first digit. There are $\color{blue}{4}$ ways to choose this first number.

Now we choose our second digit. We can't repeat numbers, so we have $3$ numbers left and therefore $\color{blue}{3}$ ways to choose the second number.

Then we choose our third digit. We can't repeat numbers, so we have $2$ numbers left and therefore $\color{blue}{2}$ ways to choose the second number.

And now we only have one number left, and so we have $\color{blue}{1}$ way to choose the last number.

So in total we have $\color{blue}{4 \times 3 \times 2 \times 1 = 24}$ ways$^{[1]}$ to create such a code.

In general, the number of ways that we can order $n$ objects (such that no object can be used twice) is $$n \times (n-1) \times (n-2) \times \cdots \times 1 = \color{blue}{n!}$$

---

$[1]$ We multiply because if there are $a$ ways to do stage 1 and $b$ ways to do stage 2, the number of ways that stage 1 and 2 can be done is $a \times b$. Likewise, for the $4$ stages we have here, we multiply the the possibility of each stage, giving $4 \times 3 \times 2 \times 1$. This is known as the rule of product.

Answer 2

Since there are no repeats, you can think of the first digit having $4$ possibilities. And because you've used that digit already, the next one has $3$ possibilities. And this continues all the way to the end. So your answer is $4\cdot3\cdot2\cdot1$ (the $1$ is there for visuals. You can exclude it and have the same product) which is just $4!$

Answer 3

First, we draw four boxes. In the first box, there are 4 possible numbers. In the second box, there are only 3 choices since numbers cannot be repeated. Repeating this process, we have $4\times3\times2\times1=24$.