Combination with repetition and no order.

Question

I'm looking for a way to calculate the number of combination of 10 choose 5 => $_5^{10}C$ while allowing any number repetitions and counting 12345 and 54321 only once (order isn't important, ie I count 11355 but not then 35115).

I think this number is majored by $10^5$, but how to remove ordering number ?

Would it be just $\left(\begin{array}{c}50\5\end{array}\right)$? — EpicGuy, Nov 22 '13 at 10:20
Does "allowing any number repetitions" mean you're including, for example, 11355 (but not then also 35153)? — aPaulT, Nov 22 '13 at 10:28
Check "combinations with repetitions" http://users.telenet.be/vdmoortel/dirk/Maths/PermVarComb.html — SJuan76, Nov 22 '13 at 10:41
@aPaulT I accept both. But I include 11355 but no then 35115 (you may make a mistake on your example) — Fractaliste, Nov 22 '13 at 10:50
@EpicGuy Are you sure ? I thought it should be lower than $10^5$ — Fractaliste, Nov 22 '13 at 10:53
@Fractaliste Well isn't just choosing 5 numbers from the set: $1,1,1,1,1,2,2,2,2,2,...,10,10,10,10,10$? — EpicGuy, Nov 22 '13 at 10:59
Quite right Fractaliste, I did mess up my example. @EpicGuy Your method, for example, counts '11112' twenty-five times (because it counts each of the ways of picking four of the five 1s in your list separately, and for each of those it then counts the five ways of picking one of the five 2s). — aPaulT, Nov 22 '13 at 11:13
@EpicGuy : it's an interesting way to understand the problem, but the result is far over the expected result. And I'm not able to see the mistake (if there is one). I'm going to think about it ! — Fractaliste, Nov 22 '13 at 11:18
Such counting of multisets is possible by the standard "stars-and-bars" method, c.f. the duplicate How to find unique multisets of n naturals of a given domain and their numbers? — hardmath, Nov 23 '13 at 16:04
@EpicGuy: this is wrong. Notice that you're counting $5,5,5,5,4$ 25 times, for instance. — tomasz, Nov 23 '13 at 20:50
@tomasz I am using combinations and I don't get how you got 25 times. — EpicGuy, Nov 25 '13 at 05:44
@EpicGuy: you have 5 copies of $5$ and 5 copies of $4$. So you can draw ${5,5,5,5,4}$ in $\binom 5 4\cdot \binom 5 1$ ways. — tomasz, Nov 25 '13 at 18:04

score 5 · Accepted Answer · answered Nov 22 '13 at 10:35

5

(Note: I'm using the notation $\binom{n}{k}$ here instead of $^n_kC$.)

A multiset is a collection of things that allows repetition but ignores order. The number of multisets of size $k$ with elements from a set of size $n$ is

$$\left(\!\!\binom{n}{k}\!\!\right) = \binom{n+k-1}{k} = \frac{(n+k-1)!}{k! (n-1)!}.$$

In your case, you'll have $\left(\!\!\binom{10}{5}\!\!\right) = \binom{10+5-1}{5} = 2002$.

answered Nov 22 '13 at 10:35

Snowball

3,078
18
37

The answer $\binom{14}{5}$ is given by the standard "stars-and-bars" method. Arrange nine "bars" (separators) and five "stars" (placeholders for elements of multisets) in a line. Any items to the left of the first bar are counted as zeros; items between the ith and i+1st bar are counted as i's; items after the ninth bar are 9's. – hardmath Nov 23 '13 at 16:11

score 1 · Answer 2 · answered Nov 22 '13 at 10:40

Since order doesn't matter, we can consider results as different if they differ after sorting. In other words, when order doesn't matter, generate the results with inherent sorting. If you build a table, 5 slots wide and 10 slots high, and trace all paths from bottom left to somewhere on the right edge, but only allow paths to go 1 step right and never go down, then a clear pattern emerges. The right hand side of this tabel contains 1,5,15,35.. or (n+3) choose 4. The sum at the right edge is 2002 or 14 ch 5. The diagonals are Pascal's triangle.

Combination with repetition and no order.

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