Math Is Fun Forum

valadezaj

Member

Registered: 2019-08-14

Posts: 2

Permutations and combinations - complex repetition.

Suppose we have the word apple. I know I can make 60 distinct permutations. However I'm not sure how to do subsets. Using "apple" suppose we want to make subsets of three. The rules are a,l and e can be used only once but p twice.
So we have [ale], [ael], [ppa] and so on. Using these rules how many distinct permutations can we make? How many combinations? Note that I already know the answer to this specific problem rather I am trying to find a generalized formula. Thanks in advance for helping!

Bob

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Registered: 2010-06-20

Posts: 10,140

Re: Permutations and combinations - complex repetition.

Hi Valadezaj,

Welcome to the forum.

If the ps can be used twice simply because there are two of them, then you may think of them as 'different' letters, let's say p1 and p2.

So you are just choosing any 3 from 5 = 5 X 4 X 3 or 5!/(5-3)!

For combinations of three letters things are a little more complicated.

If no repeated letters are chosen, calculate the number of perms and divide by the number of rearrangement s of letters, so 3x2x1/(3x2x1) = 1

If a p is chosen once, calculate similarly, so 3x2 /(2x1) = 3

If both ps are chosen we have 3 ways of choosing the third letter = 3

So 7 ways in total.

Bob

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

valadezaj

Member

Registered: 2019-08-14

Posts: 2

Re: Permutations and combinations - complex repetition.

Thanks bob for your answer but I'm not sure I explained the problem right. For the permutations the "math is fun" calculator gave me 33. The 2 p's are not suppose to be distinguishable from each other. What the calculator did was list them first - which does indeed give you 60 then removed the duplicate permutations giving me 33. But I don't know how to do that with a formula.  Hopefully this will make things clearer:

-------------
Permutations without repetition (n=5, r=3)
Using Items: a,p,p,l,e
Warning: your items have duplicates

List has 60 entries.
After removal of duplicates in result, there are now 33 entries.
{a,p,p} {a,p,l} {a,p,e} {a,l,p} {a,l,e} {a,e,p} {a,e,l} {p,a,p} {p,a,l} {p,a,e} {p,p,a} {p,p,l} {p,p,e} {p,l,a} {p,l,p} {p,l,e} {p,e,a} {p,e,p} {p,e,l} {l,a,p} {l,a,e} {l,p,a} {l,p,p} {l,p,e} {l,e,a} {l,e,p} {e,a,p} {e,a,l} {e,p,a} {e,p,p} {e,p,l} {e,l,a} {e,l,p}
------------------------

The above is how I am trying to list them. For a more generalized case I found this interesting problem on stack overflow but the explanation went a bit over my head. https://math.stackexchange.com/question … ion-subset His problem is similar to the one I posted but a bit more complex. We have the following letters: c,b,b,a,a,d and are making permutations of length 4. On that board they figured out the answer was that 102 distinct permutations could be made and indeed the "math is fun" calculator showed the same result.

--------------------------
Permutations without repetition (n=6, r=4)
Using Items: c,b,b,a,a,d
Warning: your items have duplicates

List has 360 entries.
After removal of duplicates in result, there are now 102 entries.
------------------------------------

They showed a formula on stack overflow but as I said before I didn't understand it. Hopefully now I have made things a bit clearer.

Last edited by valadezaj (2019-08-15 02:42:00)