Combinations of selecting $n$ objects with $k$ different types

Question

Suppose that I am buying cakes for a party. There are $k$ different types and I intend to buy a total of $n$ cakes. How many different combinations of cakes could I possibly bring to the party?

Answer 1

Using a method that's often called "stars and bars":

We draw $n$ stars in a row to represent the cakes, and $k-1$ bars to divide them up. All of the stars to the left of the first bar are cakes of the first type; stars between the first two bars are of the second type; . . . .

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Here's an example with $n=6$ and $k=5$. We're getting 2 of the first type, 3 of the second type, 0 of the third type, 1 of the fourth type, and 0 of the fifth type.

In order to solve the problem, we just need to reorder the stars and bars by choosing the $k-1$ spots for the bars out of the $n+k-1$ total spots, so our answer is:

$$ \binom{n+k-1}{k-1}. $$

Answer 2

Let g(n,k) = # combinations of cakes.

Notice that:

• g(n,1) = 1. (all the cakes are the same)
• g(n,2) = n+1. (e.g. for 5 cakes, the # of cakes of type 1 can be 0, 1, 2, 3, 4, 5)
• g(1,k) = k.
• g(2,k) = k*(k-1)/2 + k (the first term is two different cakes; the second term is when both cakes are the same), as long as k > 1. (otherwise g(2,1) = 1)
• g(3,k) = k * (k-1) * (k-2)/6 + k*(k-1)/2 * 2 + k (the first term is 3 different cakes; the second term is 2 different cakes, with a *2 since there are two choices for which one to duplicate, the third term is when all 3 cakes are the same), as long as k > 2.

If we think of k as a radix rather than the # of cakes, then this problem is equivalent to expressing the # of distinct n-digit numbers in base k whose digits are in sorted order. (e.g. 1122399 is equivalent to 9921231)

I think I can express it as a nonrecursive sum:

g(n,k) = sum from j=1 to max(n,k) of { (k choose j) * h(n,j) }

where h(n,j) is the # of ways to partition N cakes using j different types. (the term in the sum is when there are j distinct cakes actually chosen.)

But that's about as far as I can get... :/

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edit: looks like it's combinations with repetitions = ((k+n-1) choose n). (same as the wikipedia article with n and k swapped)

Answer 3

Let's assume you have $n$ items and $k$ bins. You need $k-1$ separators to get the $n$ items into the $k$ bins. There are $(n + k - 1)!$ permutations of ordering $n$ items and $(k-1)$ separators. The permutations of the $n$ items don't matter, and the permutations of the $(k-1)$ separators don't matter, so you'll need to divide by $n!$ and $(k-1)!$

Thus you have $$\frac{(n + k - 1)!}{n!(k-1)!} = \binom{n+k-1}{k-1}$$

Answer 4

Since cakes are involved, I shall solve it using what's called the "ice-cream parlour" approach.

For each of the $k$ types, go on saying yes, yes, yes... until you don't want any more, and then say no and move on to the next type.

Of course, you may say no straight away to any type if you don't want it, and you don't need to say no after the $k^{th}$ type because there is no next type.

So there will be $n$ of yes and $(k-1)$ of no whichever way you choose, and the only decision you need to take is where to place the no's in the string of $(n+k-1)$ responses,

thus $\dbinom{n+k-1}{k-1}$ different combinations of cakes.