Distinguishability

When distributing $n$ things to $k$ other things, one has to consider the distinguishability of the objects (i.e. if they're distinguishable or not). If the $n$ things are distinguishable, one also has to consider if duplicates are allowed (i.e. if we can repeat). For these problems, it is best to think about it first.

Distinguishable to distinguishable, with duplicates

For each of the $k$ things, there are $n$ choices, for a total of $\underbrace{n\cdot n\cdot\cdots\cdot n\cdot n}_k= n^k$ ways.

Distinguishable to distinguishable, without duplicates

For each of the $n$ things, there are $k$ choices, for a total of $\underbrace{k\cdot k\cdot\cdots\cdot k\cdot k}_n = k^n$ ways.

Distinguishable to indistinguishable, with duplicates

This is "reverse" Balls and Urns, or essentially distributing $k$ indistinguishable objects to $n$ distinguishable objects. Refer to 4; this case has $\binom{n + k - 1}{k-1}$ ways. Now we can just switch the $n$ and the $k$ , so there are $\binom{k+n-1}{n-1}$ ways.

Indistinguishable to distinguishable (Balls and Urns/Sticks and Stones/Stars and Bars)

This is the "Balls and Urns" technique. In general, if one has $n$ indistinguishable objects that one wants to distribute to $k$ distinguishable containers, then there are $\binom{n + k - 1}{k - 1}$ ways to do so.

Imagine that there are $k - 1$ dividers, denoted by $\mid$ , and $n$ objects, denoted by $\star$ , so we have $\underbrace{\mid\mid\mid\cdots\mid\mid\mid}_{k-1}$ and $\underbrace{\star\star\star\cdots\star\star\star}_n$ . Then, label the regions formed by the dividers, so we get $1\mid2\mid3\mid\cdots\mid(k - 2)\mid(k - 1)\mid k$ (since there are $k-1$ dividers) and our $n$ objects $\underbrace{\star\star\star\cdots\star\star\star}_n$ . We can now see that there are $k$ distinct regions (corresponding to the $k$ distinguishable objects) in which we can place our $n$ identical objects (corresponding to the $n$ indistinguishable objects that one is distributing), which is analogous to the original problem. Finally, there are $\binom{n+k-1}n=\binom{n + k - 1}{k - 1}$ arrangements of the $n$ stars and the $(k - 1)\ \mid\text{'s}$ by basic permutations with repeated items. Note: the number of stars that appears in each of the regions $1\mid2\mid3\mid\cdots\mid(k - 2)\mid(k - 1)\mid k$ represents the number of indistinguishable objects (the $n$ stars) given to a particular distinguishable object (of the $k$ dividers). For example, if we're distributing $9$ stars to $5$ kids, then one arrangement is $\star\mid\mid\star\star\star\mid\star\star\star\mid\star\star$ corresponding to $1$ star to the first kid, $0$ to the second, $3$ to the third, $3$ to the fourth, and $2$ to the fifth.

One problem that can be solved by this is finding the number of solutions to $a_1 + a_2 + \cdots + a_k = n$ , where $a_1,a_2,\ldots,a_n\ge0$ , which has $\binom{n + k - 1}{k - 1}$ solutions.

Indistinguishable to indistinguishable

This is part of the partition problem. Imagine that you are finding the number of solutions to $a_1 + a_2 + \cdots + a_k = n$ , where $a_1,a_2,\ldots,a_k$ are indistinguishable.

This can be done with casework; the method is best explained with an example: say that $(n,k) = (5,3)$ . Our partitions are then $\{5,0,0\},\{4,1,0\},\{3,2,0\},\{3,1,1\},\{2,2,1\}$ , so there are $5$ partitions.

This idea could also be used to solve problems like how many ways can one pay 51 cents with quarters, dimes, and pennies?

You can solve the problem as below using casework where your cases are the amounts of quarters. For each case, the amount of ways is equivalent to the amount of dimes you can use as the rest of the money can be paid by pennies.

Case 1: 2 quarters

In this case, you cannot use any dimes, leaving only 1 case.

Case 2: 1 quarter

In this case, you can have either 0, 1, or 2 dimes, leaving 3 cases.

Case 3: No quarters

In this case, you can use up to 5 dimes, leaving 6 cases.

In total we have $1+3+6=10$ cases. Note that the best way this solution can be illustrated is by a chart.

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