[Math] How many ways to put $k$ indistinguishable particles into $n$ distinguishable boxes

probabilityprobability theory

Learning probability currently through Stat$110$ Harvard's online course.

I'm confused by the Bose-Einstein problem that Blitzstein uses to convey $k$ objects out of $n$ when order doesn't matter and you don't replace.

The problem is specifically stated as follows:

How many ways to put $k$ indistinguishable particles into $n$ distinguishable boxes, where $n = 4 and k = 6?

My problem is why is $k = 6$? I thought $k$ when using "$n$ choose $k$" (even when you replace as opposed to not replacing) meant that $k$ and $n$ were the same objects.

In this problem $k$ is a totally different object: particles and boxes. In my mind "$n$ choose $k$" is for 52 cards and choose 5 from them (when order doesn't matter).

Best Answer

Just because we commonly write ${^n\mathsf C_k}$ to represent the count of ways to choose $k$ items from a set of $n$, it does not mean we always have to use $n$ and $k$ for the number of items to be chosen and size of the set, respectively.   We can count ways to choose $i$ items from a set of $j$ as: ${^i\mathsf C_j}$.   Likewise ${^{k+3}\mathsf C_{n-2}}$ represents the count of ways to choose $n-2$ items from a set of $k+3$.

Don't get hung up on the $k$, $n$, r, $\zeta$, or whatever.   They are just symbols we assign to variables in the problems.   Although we do see tendencies to use certain symbols in similar situations, there's no absolutely fixed meaning until we give them one.

The important thing is: In the selection symbol, the size of the selection source goes in the upper position, and the size of the selection goes in the lower.   Whatever symbols are used to represent their size. $${^{\lvert\text{source}\rvert}\mathsf C_{\lvert\text{selection}\rvert}}$$


In this problem there are $k$ indistinguishable particles and $n$ distinct boxes. We might represent this as $k$ stars and $n-1$ bars. Putting particals into boxes is equivalent to putting bars between stars - that is, selecting $n-1$ from $k+n-1$ spaces, placing bars there and stars in the remaining $k$ spaces.

Consider $\bbox[yellow,1pt,border:solid 0.1pt]{\star\star\star|\star\star|\star\star\star|~|\star|~}$ as representing one way to place nine particles into six boxes.

There are ${^{k+n-1}\mathsf C_{n-1}}$ distinct ways to choose $n-1$ spaces from the $k+n-1$ available.

Which also may be written as: $\dbinom {k+n-1}{n-1}$

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