# 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"?

• This is a permutation of a multiset problem since each number appears a fixed number of times. Jan 16, 2019 at 11:27

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?

• Thanks. Can you say whether these are permutations with repetition or no repetition? Jan 16, 2019 at 8:03
• I'll try. 5 distinct! objects permuted: abcde, eabcd, ..etc. certainly no repetitions. a=b, then abcde is the same as bacde(ok?), since a=b, I.e you get aacde and aacde.The same.To get rid of repetitions (overcounting) divide by 2! since a=b . Jan 16, 2019 at 8:12
• So, these are permutations with no repetition even though I have the same 4 objects (1's). Right @Peter? Jan 16, 2019 at 8:56
• Yes!!!Sorry for the roundabout way.You look at your list: 5 permutations without rep.(OK?).With 5 distinct !! objects you get 5! perm, no rep.Now let 4 be identical, you divide 5!/4!= 5(your list) to have #of perm without !! rep. Jan 16, 2019 at 9:40
• user963241Looks fine:) Jan 16, 2019 at 14:43

Yes, it is known as permutations with repetitions, and it has a formula. Say you take 5 different numbers: 1,2,3,4,5,then you can arrange them in 5!=120 days. But when you take 1,1,1,1,0, the 120 permutations will contain repeated formations. e.g. take the permutation 11110. It remains the same even when you interchange the first two 1's. In fact it remains the same even if you arrange the first four 1's in 4!=24 ways. So, 5!/4!=120/24=5 is the answer. Now you can derive the formula yourself.

• This is a permutation with a multiset since each number in the sequence appears a fixed number of times. Jan 16, 2019 at 11:29

The formula for Permutations with same repeated objects is :- (total number of letters)!/(number of repeats)!

So in your case :- 1,1,1,1,0 total numer of letters is 5 number of repeats is 4 So the answer according to formula is 5!/4!

Also to clarify things further lets see one more example :- if it was 1,1,1,1,0,0 Then it would be :- total number of letters is 6 number of repeats is 4 and 2 (1 is present 4 times and 0 is present 2 times) So according to formula the answer is 6!/(4! x 2!)

Hope this clears the concept . It is becasue if you arrange two similar objects it is the same like arranging 4 identical white balls among themselves