how many unique $3$ digit numbers can be formed using digits $1,1,1,1,2$? Explain the logic too.

Question

Only numbers $111,112,121,211$ can be formed through these. How to get this mathematically because for $1,1,1,1,1,2$ also the answer is same $111,112,121,211$ and so on as $1$'s increases answer remains same.

Answer 1

Continued from comments ...

The only possibilities are

$aaa$ for each number that we have three or more

$aab$ and permutations (whenever we have two or more $a$'s)

$abc$ and permutations (for all the digits or symbols available).

For each $aaa$ there is one permutation, for $aab$ three permutations, and for $abc$ six permutations.

Answer 2

You have plenty of $1$'s. The problem is, where to put the $2$. The $2$ could be in any of the three places, or nowhere. That gives 4 possibilities, as you've found:

111 (no twos) 112 (two in first place) 121 (two in second place) 211 (two in third place)

That's it.

Answer 3

You want to have a three digit number. You either use $2$ in which case you have three choices, namely $$211, 121,112$$ or you do not use $2$ in which case you only have one choice, namely $111$.

Thus you only have $4$ choices, even if the number of ones increase to more than three as long as you only have one $2$.