Difference between revisions of "The 2 foundational ways of counting/Statements"
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(Created page with "<noinclude> __TOC__ ==Foundations== </noinclude> Let us be given two decisions: * The first, {{M|A}}, for which we have {{M|m\in\mathbb{N}_{\ge 0} }} outcomes<ref group="Note"...") |
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If we must choose once from {{M|A}} and once from {{M|B}}, then the number of outcomes, {{M|\mathcal{O}\in\mathbb{N}_{\ge 0} }} is: | If we must choose once from {{M|A}} and once from {{M|B}}, then the number of outcomes, {{M|\mathcal{O}\in\mathbb{N}_{\ge 0} }} is: | ||
* {{M|\mathcal{O}:\eq mn}} | * {{M|\mathcal{O}:\eq mn}} | ||
| + | {{Caveat|Assume we always choose from {{M|A}} first}}, this is poorly formulated. suppose if we choose {{M|B}} first from options 1 to 6 inclusive, then we choose {{M|A}} from 7 to 12 inclusive, if we choose {{M|A}} first from 1 to 6 inclusive then we choose {{M|B}} from 7-12 inclusive then we have: | ||
| + | * {{M|\underbrace{6\times 6}_{\text{B then A} }+\underbrace{6\times 6}_{\text{A then B} } \eq 72}} ways to choose, as the set of options changes depending on whether we do {{M|B}} or {{m|A}} first - any formalism must account for this. | ||
| + | * However we must allow for some freedom, suppose we must choose two numbers from 1 to 6 inclusive, ''without'' replacement. For the first choice we can choose from 6 things, no matter what we pick for the first, the second choice is left with only 5 options. A different 5 options for each outcome of {{M|A}}, for example if {{M|A\eq 1}} then the second choice has 2 to 6 as its options, if {{M|A\eq 6}} then the second choice has 1 to 5 as its options, different from 2 to 6 | ||
===Xor=== | ===Xor=== | ||
If we must choose from either {{M|A}} or {{M|B}} - but not both, then the number of outcomes, {{M|\mathcal{O}\in\mathbb{N}_{\ge 0} }} is: | If we must choose from either {{M|A}} or {{M|B}} - but not both, then the number of outcomes, {{M|\mathcal{O}\in\mathbb{N}_{\ge 0} }} is: | ||
Revision as of 08:35, 31 August 2018
Contents
Foundations
Let us be given two decisions:
- The first, [ilmath]A[/ilmath], for which we have [ilmath]m\in\mathbb{N}_{\ge 0} [/ilmath] outcomes[Note 1]
- The second, [ilmath]B[/ilmath], for which we have [ilmath]n\in\mathbb{N}_{\ge 0} [/ilmath] outcomes.
We require that:
- There always be [ilmath]m[/ilmath] options for [ilmath]A[/ilmath] and [ilmath]n[/ilmath] options for [ilmath]B[/ilmath] irrespective of which we decide first, and irrespective of what we decide first.
- This does not mean the options cannot be different depending on which we tackle first and what we select, just that there must always be [ilmath]m[/ilmath] for [ilmath]A[/ilmath] and [ilmath]n[/ilmath] for [ilmath]B[/ilmath]
Then:
And
If we must choose once from [ilmath]A[/ilmath] and once from [ilmath]B[/ilmath], then the number of outcomes, [ilmath]\mathcal{O}\in\mathbb{N}_{\ge 0} [/ilmath] is:
- [ilmath]\mathcal{O}:\eq mn[/ilmath]
Caveat:Assume we always choose from [ilmath]A[/ilmath] first, this is poorly formulated. suppose if we choose [ilmath]B[/ilmath] first from options 1 to 6 inclusive, then we choose [ilmath]A[/ilmath] from 7 to 12 inclusive, if we choose [ilmath]A[/ilmath] first from 1 to 6 inclusive then we choose [ilmath]B[/ilmath] from 7-12 inclusive then we have:
- [ilmath]\underbrace{6\times 6}_{\text{B then A} }+\underbrace{6\times 6}_{\text{A then B} } \eq 72[/ilmath] ways to choose, as the set of options changes depending on whether we do [ilmath]B[/ilmath] or [ilmath]A[/ilmath] first - any formalism must account for this.
- However we must allow for some freedom, suppose we must choose two numbers from 1 to 6 inclusive, without replacement. For the first choice we can choose from 6 things, no matter what we pick for the first, the second choice is left with only 5 options. A different 5 options for each outcome of [ilmath]A[/ilmath], for example if [ilmath]A\eq 1[/ilmath] then the second choice has 2 to 6 as its options, if [ilmath]A\eq 6[/ilmath] then the second choice has 1 to 5 as its options, different from 2 to 6
Xor
If we must choose from either [ilmath]A[/ilmath] or [ilmath]B[/ilmath] - but not both, then the number of outcomes, [ilmath]\mathcal{O}\in\mathbb{N}_{\ge 0} [/ilmath] is:
- [ilmath]\mathcal{O}:\eq m+n[/ilmath]
Notes
- ↑ Notice we allow [ilmath]m\eq 0[/ilmath] to be, and later also for [ilmath]n\eq 0[/ilmath]. A choice with nothing to choose from is not a decision at all, so zero has meaning
References
