Permutations with same repeated objects Suppose I have these numbers: $$1,1,1,1,0.$$ When I arrange them to find permutations I will get: $01111$ $10111$ $11011$ $11101$ $11110$ Is this is the idea what is commonly known as "Permutations with repetition"? If so, then I am not getting the excepted answer: e.g. n! = 5! = 120 permutations. E.g: $$n!/(n-r)! = 120/(5-5)! = 120$$ I am not getting an excepted answer. Is this case something different? Is there a formula? Is what I am trying to do known as "combinations with repetition"?

Going backward:

Take any one of your permutations, say

$11110.$

Assume you had $a,b,c,d$ distinct digits ($\
ot =0$)

Look at $abcd0$. The number of distinct permutations with the $0$ fixed in the last slot is:

$4!$ .

Now you have $5$ permutations listed with four $1$'s and one $0$:

Look at

$abcd0, abc0d,ab0cd, a0bcd, 0abcd.$

where $a,b,c,d$ are distinct ($\
ot =0$).

You get $5 × 4!=5!$ distinct permutations (Why?).

Finally :

If you have five objects to permute where $4$ are identical, you get:

$5!/4!=5$ distinct permutations .

Can you generalize for $n$ objects , $r$ of which are identical? How many distinct permutations do you get?