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"?
Best Answer
Going backward:
Take any one of your permutations, say
$11110.$
Assume you had $a,b,c,d$ distinct digits ($\not =0$)
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 ($\not =0$).
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?