Difference between revisions of "Distinguishability"
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==Distinguishable to distinguishable, with duplicates== | ==Distinguishable to distinguishable, with duplicates== | ||
For each of the <math>k</math> things, there are <math>n</math> choices, for a total of <math>\underbrace{n\cdot n\cdot\cdots\cdot n\cdot n}_k= n^k</math> ways. | For each of the <math>k</math> things, there are <math>n</math> choices, for a total of <math>\underbrace{n\cdot n\cdot\cdots\cdot n\cdot n}_k= n^k</math> ways. | ||
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==Distinguishable to distinguishable, without duplicates== | ==Distinguishable to distinguishable, without duplicates== | ||
For each of the <math>n</math> things, there are <math>k</math> choices, for a total of <math>\underbrace{k\cdot k\cdot\cdots\cdot k\cdot k}_n = k^n</math> ways. | For each of the <math>n</math> things, there are <math>k</math> choices, for a total of <math>\underbrace{k\cdot k\cdot\cdots\cdot k\cdot k}_n = k^n</math> ways. | ||
+ | |||
==Distinguishable to indistinguishable, with duplicates== | ==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 | + | This is "reverse" Balls and Urns, or essentially distributing <math>k</math> indistinguishable objects to <math>n</math> distinguishable objects. Refer to 4; this case has <math>\binom{n + k - 1}{k-1}</math> ways. Now we can just switch the <math>n</math> and the <math>k</math>, so there are <math>\binom{k+n-1}{n-1}</math> ways. |
− | == | + | ==Indistinguishable to distinguishable ([[Stars and bars|Balls and Urns]]/[[Stars and bars|Sticks and Stones]]/[[Stars and bars|Stars and Bars]])== |
− | This is | + | This is the "Balls and Urns" technique. In general, if one has <math>n</math> indistinguishable objects that one wants to distribute to <math>k</math> distinguishable containers, then there are <math>\binom{n + k - 1}{k - 1}</math> ways to do so. |
− | + | Imagine that there are <math>k - 1</math> dividers, denoted by <math>\mid</math>, and <math>n</math> objects, denoted by <math>\star</math>, so we have <math>\underbrace{\mid\mid\mid\cdots\mid\mid\mid}_{k-1}</math> and <math>\underbrace{\star\star\star\cdots\star\star\star}_n</math>. Then, label the regions formed by the dividers, so we get <math>1\mid2\mid3\mid\cdots\mid(k - 2)\mid(k - 1)\mid k</math> (since there are <math>k-1</math> dividers) and our <math>n</math> objects <math>\underbrace{\star\star\star\cdots\star\star\star}_n</math>. We can now see that there are <math>k</math> distinct regions (corresponding to the <math>k</math> distinguishable objects) in which we can place our <math>n</math> identical objects (corresponding to the <math>n</math> indistinguishable objects that one is distributing), which is analogous to the original problem. Finally, there are <math>\binom{n+k-1}n=\binom{n + k - 1}{k - 1}</math> arrangements of the <math>n</math> stars and the <math>(k - 1)\ \mid\text{'s}</math> by basic permutations with repeated items. '''Note:''' the number of stars that appears in each of the regions <math>1\mid2\mid3\mid\cdots\mid(k - 2)\mid(k - 1)\mid k</math> represents the number of indistinguishable objects (the <math>n</math> stars) given to a particular distinguishable object (of the <math>k</math> dividers). For example, if we're distributing <math>9</math> stars to <math>5</math> kids, then one arrangement is <math>\star\mid\mid\star\star\star\mid\star\star\star\mid\star\star</math> corresponding to <math>1</math> star to the first kid, <math>0</math> to the second, <math>3</math> to the third, <math>3</math> to the fourth, and <math>2</math> to the fifth. | |
− | <math>\ | ||
− | <math>\{ | ||
− | <math>\{ | ||
− | <math>\{ | ||
− | <math>\ | ||
− | + | One problem that can be solved by this is finding the number of solutions to <math>a_1 + a_2 + \cdots + a_k = n</math>, where <math>a_1,a_2,\ldots,a_n\ge0</math>, which has <math>\binom{n + k - 1}{k - 1}</math> solutions. | |
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==Indistinguishable to indistinguishable== | ==Indistinguishable to indistinguishable== | ||
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This can be done with casework; the method is best explained with an example: say that <math>(n,k) = (5,3)</math>. Our partitions are then <math>\{5,0,0\},\{4,1,0\},\{3,2,0\},\{3,1,1\},\{2,2,1\}</math>, so there are <math>5</math> partitions. | This can be done with casework; the method is best explained with an example: say that <math>(n,k) = (5,3)</math>. Our partitions are then <math>\{5,0,0\},\{4,1,0\},\{3,2,0\},\{3,1,1\},\{2,2,1\}</math>, so there are <math>5</math> partitions. | ||
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− | |||
− | + | 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 <math>1+3+6=10</math> cases. Note that the best way this solution can be illustrated is by a chart. | ||
[[Category:Combinatorics]] | [[Category:Combinatorics]] |
Latest revision as of 11:36, 31 July 2024
When distributing things to
other things, one has to consider the distinguishability of the objects (i.e. if they're distinguishable or not). If the
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.
Contents
Distinguishable to distinguishable, with duplicates
For each of the things, there are
choices, for a total of
ways.
Distinguishable to distinguishable, without duplicates
For each of the things, there are
choices, for a total of
ways.
Distinguishable to indistinguishable, with duplicates
This is "reverse" Balls and Urns, or essentially distributing indistinguishable objects to
distinguishable objects. Refer to 4; this case has
ways. Now we can just switch the
and the
, so there are
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 indistinguishable objects that one wants to distribute to
distinguishable containers, then there are
ways to do so.
Imagine that there are dividers, denoted by
, and
objects, denoted by
, so we have
and
. Then, label the regions formed by the dividers, so we get
(since there are
dividers) and our
objects
. We can now see that there are
distinct regions (corresponding to the
distinguishable objects) in which we can place our
identical objects (corresponding to the
indistinguishable objects that one is distributing), which is analogous to the original problem. Finally, there are
arrangements of the
stars and the
by basic permutations with repeated items. Note: the number of stars that appears in each of the regions
represents the number of indistinguishable objects (the
stars) given to a particular distinguishable object (of the
dividers). For example, if we're distributing
stars to
kids, then one arrangement is
corresponding to
star to the first kid,
to the second,
to the third,
to the fourth, and
to the fifth.
One problem that can be solved by this is finding the number of solutions to , where
, which has
solutions.
Indistinguishable to indistinguishable
This is part of the partition problem. Imagine that you are finding the number of solutions to , where
are indistinguishable.
This can be done with casework; the method is best explained with an example: say that . Our partitions are then
, so there are
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 cases. Note that the best way this solution can be illustrated is by a chart.