How many 4 digit numbers can be formed by digits 3, 1, 3, 1?

Question

This a question from my textbook and it says the answer is 6 and I fail to see how. Per my understanding if we consider each digit being used only once i.e viewing all 4 digits as distinct elements (and not two 1s and two 3s), we get the answer 4 factorial i.e 24.

Answer 1

The issue is that not every arrangement of the four digits is unique.

For example, there are multiple ways to form $1133$ - you could swap the $1$'s, the $3$'s, or both. Different rearrangements (in a sense), but the number is the same.

Thus, you need to account for that repetition. The way to handle it would be to divide by the factorials of each digit repeats. For example, we have two $3$'s, so we divide by $2!$. Since we have two $1$'s, similarly, we divide by that again.

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Related example to cement this fact:

Consider the number of 8-digit numbers we can make from $9, 9, 9, 9, 7, 7, 7, 2$.

We first consider if these were unique: there are 8 digits, so the number of unique arrangements would be $8!$.

But we notice we have repeated digits. You could swap them around in the number you get, but the number would be same. You divide, then, by the number of ways you could rearrange those repeated digits (thus cutting out the repetitions).

We have $4$ $9$'s, so we will divide by $4!$. We have $3$ $7$'s, so we divide again by $3!$. We also have $1$ $2$ - we could divide by $1!$, but since $1! = 1$, we don't have to worry about that.

Thus in total, there are

$$\frac{8!}{4! \cdot 3!}$$

rearrangements.

Answer 2

If the 4 digits are distinct then you'll have $4!=24$ ways to to form the number.

But since there is 2 $1$'s, we divide by $2!$

And since there is 2 $3$'s, we divide another time by $2!$

So the number of ways to form the number is $\frac{24}{2!.2!} =6$

Note that we divide by $2!$ to avoid the repetition of the same number twice i.e. 1133 will stay the same if we swap the $1$'s. So by dividing by $2!$ we will be counting the number $1133$ once and not twice.

For example when you have 3 $1$'s you divide the total number of ways by $3!$ to avoid the repetition of the same number just by swapping the $1$'s