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I'm a non-math major but have to study a course in probability and statistics for graduate school. I was able to understand permutations with repetition and then permutations without repetition using examples, and then trying to generalize.

But I can't seem to find any examples of combinations with repetition that explain how the formula works, without going into multisets and things like that.

Help would be much appreciated.

N. F. Taussig
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RadDragon
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3 Answers3

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one practical example is the following:

Suppose that you and 7 other friends go to a restaurant.

You each want to order a sandwich, and the restaurant offers 3 varieties of sandwiches (A, B and C). Then combinations with repetitions answer the question "How many different orders can you make"? It's not important who orders which sandwich, only the amount of sandwiches of each type ordered.

You can represent each sandwich to be ordered with an X. So you need to count the ways to distribute 8 X's in three groups:

X X X X X X X X

You can add 2 separator marks to distribute the X's in three subsets, and then claim that the first space is for sandwich of type A, the second for sandwich of type B and the third for sandwiches of type C. Some examples of these are

X X X | X X X X | X (3 A sandwiches, 4 B sandwiches, 1 C sandwich)

X X | X X | X X X X (2 A sandwiches, 2 B sandwiches, 4 C sandwich)

| X X X X X X | X X (0 A sandwiches, 6 B sandwiches, 2 C sandwich)

Note that this represents every way to distribute the sandwiches, and thus counting how to arrange the 2 | symbols and the 8 X gives the solution to the problem.

So, if $CR^3_8$ denotes the combinations with repetition that solve this problem, then

$$CR^3_8 = {3 + 8 - 1\choose8}$$

Because that's the ways you can rearrange the X's and the |'s

Jean-Claude Arbaut
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Joaquín Lejtreger
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Combinations with repetition is choosing $k$ objects from a basket with $n$ distinct objects that magically replaces every item you choose so that your choices don't reduce. (Only your $k$th choice isn't replaced as the process gets completed)

In effect, we have added an extra $(k-1)$ items to choose from, hence the formula is

$$\dbinom{n+k-1}{k} $$

true blue anil
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    I wish my professor was this good. Thank you – RadDragon Sep 12 '21 at 10:14
  • Glad to have been of help, happy that you liked it ! – true blue anil Sep 12 '21 at 11:56
  • @trueblueanil I just saw this as you referenced it in a comment to a solution you wrote recently. Could you expand a little on how "In effect, we have added an extra $(k−1)$ items to choose from" leads to the upper index being $n+k-1$? I'm drawing a blank on how to connect the two. (I am familiar with the stars-and-bars argument.) – A.J. Jun 02 '23 at 06:13
  • @A.J. Sorry, have just seen your query. $n$ items to choose from with which we started expands to $n+k-1$ items because each time we are choosing an item of some type, we are "magically" adding $1$ item of the same type except for the last choice. – true blue anil Jun 04 '23 at 20:38
  • @A.J. The start should read "one each of $n$ types of items to choose from..." – true blue anil Jun 04 '23 at 21:01
  • @trueblueanil Thanks for replying. What I'm missing is how just adding $k-1$ to the upper index accounts for all the variations of what those $k$ items might be; from all $k$ being of a single type, to each being one of $k$ different types, and all the possibilities in between. If this is just an intuitive argument I'll happily accept that; but if this line of thought could be fleshed out into a more rigorous argument, I'd be grateful for any pointers on how to do that. – A.J. Jun 06 '23 at 03:59
  • @A.J. For each of $k$ choices, I always have the full basket of $n$ types of objects to choose from, and I am free to choose any type each time. I am not a mathematician, so if this is not "rigorous" , I don't know how to make it so. The other answer has used a stars and bars argument which is what I was trying to escape from to give the argument a different flavour. – true blue anil Jun 06 '23 at 04:24
  • Got it, thanks. – A.J. Jun 06 '23 at 04:30
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I frame combinations-with-repetition problems as choosing/picking (order doesn't matter) $r$ items from $n$ categories (each category may be repeatedly picked).

For each selection, illustrate the $n$ categories using $(n-1)$ vertical bars as separators, then represent the $r$ selected items by filling $r$ stars in the according categories.

Notice that every selection contains $[(n-1)+r]$ positions filled by stars & bars, and that selections are distinguished by the $r$ stars filling in different combinations of positions (the vertical bars filling in the remaining positions).

Therefore, the total number of selections is the number of ways to choose $r$ positions from $[(n-1)+r]$ positions, i.e., $$\binom{n-1+r}{r}.$$

ryang
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