If 3 things can be in one of 2 states,
the number of possible combinations
is equal to 2^3.
But if I have 3 things, 2 of which can
be in 2 states and the other in 3 states,
what's the simplest expression that will
accurately compute the permutationjs
possible? 22 4935
That's a bad example because it uses the same number twice, so this
might not make sense but the best formula to compute that would be 2X +
3Y, where X = 2 and Y = 1. X + Y = the number of things you're talking
about.
MLH wrote:
If 3 things can be in one of 2 states,
the number of possible combinations
is equal to 2^3.
But if I have 3 things, 2 of which can
be in 2 states and the other in 3 states,
what's the simplest expression that will
accurately compute the permutationjs
possible?
ManningFan <ma********@gma il.comwrote:
: That's a bad example because it uses the same number twice, so this
: might not make sense but the best formula to compute that would be 2X +
: 3Y, where X = 2 and Y = 1. X + Y = the number of things you're talking
: about.
Are you assuming 2 'things' capable of being in 2 states
simultaneously? I assumed that MLH meant
But if I have 3 things, 2 of which can
be in *1 of* 2 states and the other
in *1 of* 3 states:
2*2*3 ot #possPos1 * #possPos2 * #possPos1 3
My other, less favored, interpretation would be that
things 1 and 2 form a joint thing that can be in 2
states, e.g. state 1: thing 1=red, thing 2=blue
state 2: thing 1=white, thing 2=brown
2*3 or #possjointPos1, 2 * #possPos3
for poss[ibilities] and Pos[itions]
thelma
:>
: MLH wrote:
:If 3 things can be in one of 2 states,
:the number of possible combinations
:is equal to 2^3.
:>
:But if I have 3 things, 2 of which can
:be in 2 states and the other in 3 states,
:what's the simplest expression that will
:accurately compute the permutationjs
:possible?
MLH wrote:
If 3 things can be in one of 2 states,
the number of possible combinations
is equal to 2^3.
But if I have 3 things, 2 of which can
be in 2 states and the other in 3 states,
what's the simplest expression that will
accurately compute the permutationjs
possible?
(2^2)*3

Rick Brandt, Microsoft Access MVP
Email (as appropriate) to...
RBrandt at Hunter dot com
Would like to thank all who contributed.
Rick, your answer looks the simplest. To
be sure, though, I'll add clarity to the Q
for Manning's and Thelma's benefit...
You walk into a room which has nothing
in it but some water, a lite bulb and a radio.
If the water can be in 3states: solid, liquid or gas &
If the lite can be on or off &
if the radio can be on or off
How many conditions (each condition being
a single unique combination of states of the
above 3 objects) are possible?
According to Rick, 12 ==2 to the 2nd power,
all that times 3. I agree. A brute force approach
to the question proves his answer correct, I think.
Even though, as an answer, it is correct  its not
the algorithm I was seeking. However, to Rick's
credit, an algorithm can be inferred I think. Perhaps
its this:
If you have a collection of any number of things
that can be in any one of any number of states,
they can be grouped so that all things that can be
in the same number of states together. Call these
groups 'factors'. Make a exponential calculation
in each group, raising the number of states (base)
to the power of the number of things (exponent).
Then multiply the factors. The answer equals the
permutations.
In the water, lite bulb, radio example; there are two
things that can be in two states (radio & lite bulb)
and one thing that can be in two states (water). So
the radio & lite bulb are grouped together in the
group of things that can be in 2states. The water
is in a group by itself that can be in 3states. Now
the factors:
First group: 2 states raised to the power of 2 things = 4
Sec group: 3 states raised to the power of 1 thing = 3
Multiplying your factors, 3 x 4 gives you 12.
Sound about rght, Rick? Anyone else agree? Disagree?
Rick, if that's right, and I apply the algorithm to the following
dilemma: I have 8 marbles that can be red, blue or green,
2 frogs that can be jumping or not, 3 people that can be
eating, sleeping or scuba diving and 4outlaws that can
be dead or alive. All these things are out on the beach at
a swell dive spot. The number of possible combinations
of all these things (permutations) is 177183. Is that what
you would come up with, Rick? Note: I'm not certain of
this answer. Reverse engineering your answer to imply an
algorithm and using it is all I did here.
Excuse me. I made a computational error. The number
I meant to post was 6,377,292 == not 177,183.
"MLH" <CR**@NorthStat e.netwrote in message
news:1n******** *************** *********@4ax.c om...
Would like to thank all who contributed.
Rick, your answer looks the simplest. To
be sure, though, I'll add clarity to the Q
for Manning's and Thelma's benefit...
You walk into a room which has nothing
in it but some water, a lite bulb and a radio.
If the water can be in 3states: solid, liquid or gas &
If the lite can be on or off &
if the radio can be on or off
How many conditions (each condition being
a single unique combination of states of the
above 3 objects) are possible?
According to Rick, 12 ==2 to the 2nd power,
all that times 3. I agree. A brute force approach
to the question proves his answer correct, I think.
Even though, as an answer, it is correct  its not
the algorithm I was seeking. However, to Rick's
credit, an algorithm can be inferred I think. Perhaps
its this:
If you have a collection of any number of things
that can be in any one of any number of states,
they can be grouped so that all things that can be
in the same number of states together. Call these
groups 'factors'. Make a exponential calculation
in each group, raising the number of states (base)
to the power of the number of things (exponent).
Then multiply the factors. The answer equals the
permutations.
Don't over think this. Possible combinations are derived by multiplying the
possible states of each entity by the possible states of the others. Raising to
a power is merely shorthand for the multiplying and is only possible any time
the number of possible states between entities is the same. It is not when the
possible number of states is different.
A die has 6 possible states so 2 dice have 6^2 outcomes and 3 dice have 6^3, but
only when all die have the same number of sides. If I have one die with 6 sides
and one with 5 then there are 5*6 possible outcomes.

Rick Brandt, Microsoft Access MVP
Email (as appropriate) to...
RBrandt at Hunter dot com
On Sat, 16 Dec 2006 16:14:08 GMT, "Rick Brandt"
<ri*********@ho tmail.comwrote:
>Don't over think this. Possible combinations are derived by multiplying the possible states of each entity by the possible states of the others. Raising to a power is merely shorthand for the multiplying and is only possible any time the number of possible states between entities is the same.
Gotcha. So, applying the above to the question regarding the marbles,
frogs, people and outlaws, whats the number of combinations you come
up with?
MLH <CR**@NorthStat e.netwrote:
: On Sat, 16 Dec 2006 16:14:08 GMT, "Rick Brandt"
: <ri*********@ho tmail.comwrote:
:>Don't over think this. Possible combinations are derived by multiplying the
:>possible states of each entity by the possible states of the others. Raising to
:>a power is merely shorthand for the multiplying and is only possible any time
:>the number of possible states between entities is the same.
: Gotcha. So, applying the above to the question regarding the marbles,
: frogs, people and outlaws, whats the number of combinations you come
: up with?
Using basic I got that
3^8 * 2^2 * 3^3 * 2^4 = 11337408
marbles frogs people outlaws
I think that your algorithm with its shortcut is fine. In aircode
[even worse than aircode I'm afraid] form:
if N=number of EntityTypes()= array of number of Entities of each Type,
EntityStates()= array of number of States for each EntityType
Combinations = 1
for Type = 1 to N
Combinations = Combinations *
EntityStates(Ty pe)^EntityTypes (Type)
thelma
So we got two different answers. That's what I was afraid of.
I haven't a clue who's right. I'm not sure what algorithm Rick was
using. Thx for the input. This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics 
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