# How many 3-digit numbers can be formed with the digits $1, 2, 3, 4$?

how many $$3$$ digit numbers can be formed by $$1,2,3,4$$, when the repetition of digits is allowed?

So basically, I attempted this question as-

There are 4 numbers and 3 places to put in the numbers: In the ones place, any 4 numbers can be put, so there are 4 choices in the ones place. Similarly for the tens and the hundreds place. So, the total choices are, by multiplication principle- $$4*4*4=64$$ And well and good, this was the answer.

But what if I reversed the method?

So I take some particular numbers, like $$1,2,3$$ and say that, well, $$1$$ can go in $$3$$ places, $$2$$ in $$2$$ places and $$3$$ in $$1$$ place, so by multiplication principle, there are $$6$$ ways of forming a $$3$$-digit number with $$1,2,3$$.

But there are $$4$$ different numbers. So the number of $$3$$-number combinations are- $$(1,2,3)$$,$$(1,2,4)$$,$$(1,3,4)$$,$$(2,3,4)$$. Each can be arranged in $$6$$ ways, so we get $$24$$ ways totally.

So why is my answer different here?

## 3 Answers

Your answer is different because in the question, repitition is allowed, but you have only chosen combinations $$(1,2,3),(1,2,4),(1,3,4),(2,3,4)$$ in which there are no repeat numbers. So the combinations that you were supposed to include were $$(1,1,1),(2,2,2),(3,3,3),(4,4,4), (1,2,2), (1,3,3),(1,4,4)...$$ and so on.

Now, if you count the permutations for each of these combinations separately, and add it to 24, you will get 64.

Hope this helps :)

• So, my method was correct? Feb 21, 2020 at 6:09
• Yes, your method would have been correct if you remembered to take combinations of repeated numbers. Unfortunately, it is a not easy to simply "know" when a method is correct. That requires practice and effort. Mathematics is all about practice. Feb 21, 2020 at 6:18

Well I think this is because when you do $$4*4*4$$ , you consider that the digits in the number can repeat themselves. But in the second case , you take $$3$$ cases in first place , $$2$$ in second and $$1$$ in third. So then you are considering numbers with all distinct digits..

Example- In first method- A number can be $$111$$ But in second case - A number cannot be $$111$$ , it will be like $$123$$ or $$124$$

That's why I think some numbers will not be counted....

The answer is 64.

Imagine there are 3 empty spaces and for each place we have 4 numbers (or 4 choices) to bill the place with. So, we have 4×4×4 ways in all

• Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. Jul 24, 2023 at 13:07