Difference between revisions of "The 6 primitives of counting"

Latest revision as of 08:46, 31 August 2018

Foundations

Let us be given two decisions:

• The first, $A$, for which we have $m\in\mathbb{N}_{\ge 0}$ outcomes^{[Note 1]}
• The second, $B$, for which we have $n\in\mathbb{N}_{\ge 0}$ outcomes.

We require that:

• There always be $m$ options for $A$ and $n$ options for $B$ 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 $m$ for $A$ and $n$ for $B$

Then:

And

If we must choose once from $A$ and once from $B$, then the number of outcomes, $\mathcal{O}\in\mathbb{N}_{\ge 0}$ is:

• $\mathcal{O}:\eq mn$

Caveat:Assume we always choose from $A$ first, this is poorly formulated. suppose if we choose $B$ first from options 1 to 6 inclusive, then we choose $A$ from 7 to 12 inclusive, if we choose $A$ first from 1 to 6 inclusive then we choose $B$ from 7-12 inclusive then we have:

• $\underbrace{6\times 6}_{\text{B then A} }+\underbrace{6\times 6}_{\text{A then B} } \eq 72$ ways to choose, as the set of options changes depending on whether we do $B$ or $A$ 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 $A$, for example if $A\eq 1$ then the second choice has 2 to 6 as its options, if $A\eq 6$ then the second choice has 1 to 5 as its options, different from 2 to 6

Xor

If we must choose from either $A$ or $B$ - but not both, then the number of outcomes, $\mathcal{O}\in\mathbb{N}_{\ge 0}$ is:

• $\mathcal{O}:\eq m+n$
• Caveat:Consider the case where: $A$ is a choice from 1 to 5 inclusive, and $B$ is a choice from 5 to 10 (note the 6 outcomes, 5,6,7,8,9 or 10) inclusive, if we use the disjoint union for the set of outcomes, we get 11 outcomes, which correspond to $(A,1),\ldots,(A,5),(B,5),(B,6),\ldots,(B,10)$, if we use just the union then the outcomes are 1,2,3,4,5,6,7,8,9,10 - 10 outcomes. So again this needs to be defined better.

Primitives

The 4 primitive ways of counting/Statements

Notes

1. Jump up ↑ Notice we allow $m\eq 0$ to be, and later also for $n\eq 0$. A choice with nothing to choose from is not a decision at all, so zero has meaning

References