Difference between revisions of "Distinguishability"

Revision as of 22:14, 2 March 2009

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.

1 Distinguishable to distinguishable, with duplicates
2 Distinguishable to distinguishable, without duplicates
3 Distinguishable to indistinguishable, with duplicates
4 Distinguishable to indistinguishable, without duplicates
5 Indistinguishable to indistinguishable
6 Indistinguishable to distinguishable (Balls and Urns)

Distinguishable to distinguishable, with duplicates

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

Distinguishable to distinguishable, without duplicates

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

Distinguishable to indistinguishable, with duplicates

This is "reverse" Balls and Urns, or essentially distributing $k$ indistinguishable objects to $n$ distinguishable objects. Refer to 6; this case has $\binom{n + k - 1}k$ ways (notice the difference between this and 6). This works because in every distribution, since the number of indistinguishable objects $k$ is fixed, it only matters how many times each of the $n$ distinguishable objects were used. If we let the number of times each of the $n$ distinguishable objects were used be nonnegative $a_1,a_2,\ldots,a_n$ , then we must have $a_1+a_2+\cdots+a_n=k$ , which, as seen in 6, has $\binom{k+n-1}{n-1}=\binom{n+k-1}k$ solutions.

Distinguishable to indistinguishable, without duplicates

This is probably the most tedious cases, as it involves the most casework. One way is to first find all the partitions (refer to 5) of $n$ with $k$ addends, (i.e. all solutions to $a_1 + a_2 + \cdots + a_k = n$ in which the addends are indistinguishable). Then, for each partition, separately calculate the number of ways, and finally, add these results together.

For example, if $(n,k) = (5,3)$ , then our partitions are: $\{5,0,0\}$ --this case has $1$ way. $\{4,1,0\}$ --we choose one of $n$ to be the " $1$ ", so there are $5$ ways. $\{3,2,0\}$ --we choose three objects to be the " $3$ 's" (the rest are determined after this), so there are $\binom53 = 10$ ways for this. $\{3,1,1\}$ --again, we choose three objects to be the " $3$ 's" (the rest are determined after this), so there are $\binom53 = 10$ ways for this. $\{2,2,1\}$ --first, we choose one object to be the " $1$ ", which has $5$ ways. Then, we can choose any two of the remaining four to be one of the " $2$ 's", and there are $\binom42 = 6$ ways for this. However, we must divide this by $2$ , since the two " $2$ 's" are interchangeable, and the total for this case is $5\cdot6\cdot\frac12 = 15$ .

Adding up, we get $1 + 5 + 10 + 10 + 15 = 41$ ways.

All of these problems are similar to this one in that you divide them up into smaller counting problems.

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.

Indistinguishable to distinguishable (Balls and Urns)

This is "Balls and Urns". In general, if one has $n$ indistinguishable objects that one wants to distribute to $k$ distinguishable objects, 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 $(k - 1)\ \mid\text{'s}$ , which is our desired amount. Note that 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).

One common 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.

@@ Line 5: / Line 5: @@
 For each of the <math>n</math> things, there are <math>k</math> choices, for a total of <math>\underbrace{k\cdot k\cdots k\cdot k}_n = k^n</math> ways.
 ==Distinguishable to indistinguishable, with duplicates==
-This is "reverse" Balls and Urns, or essentially distributing <math>k</math> indistinguishable objects to <math>n</math> distinguishable objects. Refer to 6; this case has <math>\binom{n + k - 1}k</math> ways (notice the difference between this and 6).
+This is "reverse" Balls and Urns, or essentially distributing <math>k</math> indistinguishable objects to <math>n</math> distinguishable objects. Refer to 6; this case has <math>\binom{n + k - 1}k</math> ways (notice the difference between this and 6). This works because in every distribution, since the number of indistinguishable objects <math>k</math> is fixed, it only matters how many times each of the <math>n</math> distinguishable objects were used. If we let the number of times each of the <math>n</math> distinguishable objects were used be nonnegative <math>a_1,a_2,\ldots,a_n</math>, then we must have <math>a_1+a_2+\cdots+a_n=k</math>, which, as seen in 6, has <math>\binom{k+n-1}{n-1}=\binom{n+k-1}k</math> solutions.
 ==Distinguishable to indistinguishable, without duplicates==
 This is probably the most tedious cases, as it involves the most casework. One way is to first find all the partitions (refer to 5) of <math>n</math> with <math>k</math> addends, (i.e. all solutions to <math>a_1 + a_2 + \cdots + a_k = n</math> in which the addends are indistinguishable). Then, for each partition, separately calculate the number of ways, and finally, add these results together.

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