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.com

  • 2
    $\begingroup$ Hint : How many possible positions are there for the ones ? $\endgroup$
    – Peter
    Dec 15, 2018 at 8:45

2 Answers 2


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.

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

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


  • $\begingroup$ So, is the correct approach to this 4 choose 2? $\endgroup$ Dec 15, 2018 at 8:49
  • $\begingroup$ Not quite - that the same answer comes out is a mere coincidence. Notice how that doesn't generalize if, for a different example, we were told to find the number of 6-digit numbers for $4, 3, 3, 2, 2, 2$. What the correct/intended method is, I wouldn't know since I haven't taken a combinatorics course. My method would be just to notice that, for each individual number, you have three superfluous numbers: you can swap the $1$'s for one of them, swap the $3$'s for another, and swap both pairs for a third. $\endgroup$ Dec 15, 2018 at 8:54
  • $\begingroup$ So then you notice that each "unique" number has three duplicates as well, i.e. $3/4$ of the permutations (were the digits unique) are pointless. Or more properly, only $1/4$ of the numbers you would get by permuting them as if they were unique are actually unique. Thus, the number of unique numbers would be $$\frac{1}{4} \cdot 4!$$ $\endgroup$ Dec 15, 2018 at 8:54
  • $\begingroup$ Granted that's just how I personally would handle it. How a textbook/classroom setting would expect you to handle it, I don't know, owing to not having been in those environments. $\endgroup$ Dec 15, 2018 at 8:55
  • $\begingroup$ Looking at the other answer, the proper way would be to divide by the number of repeated digits. For example, for our 6-digit problem earlier, the answer would be $$\frac{6!}{2! \cdot 3!}$$ Or for the number of 10-digit numbers made by $1, 2, 2, 3, 3, 3, 4, 4, 4, 4$: $$\frac{10!}{2! \cdot 3! \cdot 4!}$$ because we have two $2$'s, three $3$'s, and four $4$'s. $\endgroup$ Dec 15, 2018 at 8:58

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


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