Combinations with limited repitition

[soandos]

How can one find the number of combinations of a list such as {2,2,2,2,2,2,2,3,3,3,3,5,7} (7 2s, 4 3s, 1 5, 1 7). I am pretty sure that the answer is 160 (the number of proper divisors of 9!). How can I figure it out using combinations/permutations or some other method?

 

[Karlx]

Hi Soandos.

You have a list with 13 members, in groups of 7,4,1 and 1.
Let us proceed step by step.
1) In how may ways may you combine 13 members ?
2) Among all these ways, how will you take into account the groups of equals members, of size 7,4,1 and 1?

 

[soandos]

sum of binomials = 8192
dividing by 7! then 4! does not work...
Am I correct in assuming that 160 is the correct answer?

 

[soandos]

is the fact that its 2^13 coincidence?

 

[Karlx]

Suppose you have the following list: {a,b,c}.
In how many ways may you arrange these three members ?

 

[soandos]

Combinations?
a,b,c
a,b
b,c
a,c
a
b
c
0

so 8 = 3!/1! + 2(the null and the full) (1 being number of repeated elements)
doing that for my original set gives 158 if i don't differentiate between what gets repeated (i.e. 13!/12! + 2 = 158) but it seems that there are more.

 

[uart]

There is only one combination that uses all 13 elements so I assume that's not your question.

Instead of me guessing any more please define your question properly.

- Do you want permutations using all 13 elements?
- Combinations that use one each of the four different elements?
- etc etc. There really are a lot of DIFFERENT questions that I could make from that list!

Don't make people guess what the question is, specify it fully!

 

[soandos]

I want ALL possible combinations, of every length, using only the elements in the list. 2,2,3 is indistinct from 2,3,2.

 

[soandos]

to restate the question differently (I hope these statements are equivalent) given the prime factors of a number and their powers, how many proper divisors does it have?

 

[Karlx]

So, if I understand you, you have the number

362880 = 2^{7}\cdot3^{4}\cdot5\cdot7

and you want to calculate how may proper divisors does it have.

If one of the factors has exponent n, then, counting also the zero exponent, this factor may arise in n+1 ways. So for the other factors.

Thus the number of proper divisors is (7+1)\cdot(4+1)\cdot2\cdot2=160

From a combinatorial point of view, I suppose you can read this as follows:

Number 2 appears 7 times in the initial set.
So in the possible combinations, it can appear in (7+1) possible ways (one for the number 2 not appearing).
In the same way, number 3 can appear in (4+1) possible ways.
And the same for number 5 and number 7.
So the number of combinations would be the product.

 

[soandos]

thank you.