Math help

[DonW]

Sad but six months ago I new how to do this now its gone, I blame Boomer
Take a look, this is for my heaters. Help or a reminder would be great.

Three sets of numbers

1,2 Group 1
3,4 Group 2
5,6,7 Group 3

I need to know the total possible combinations with one catch.

Example:
1,3,5 only one number from each group.
or
1,4,6
or
1,3,7

Thanks
Don

 

[LakeEd]

Don,

I may not be totally following your question... but you have 2 numbers in group 1... 2 numbers in group 2, and 3 numbers in group 3

2 X 2 X 3 = 12 combinations... with 1 number from each group.

 

[DonW]

Don,

I may not be totally following your question... but you have 2 numbers in group 1... 2 numbers in group 2, and 3 numbers in group 3

2 X 2 X 3 = 12 combinations... with 1 number from each group.

Its alot more than 12. There's an equation, it fell out of my overflow.

Don

 

[Pikachu]

12 is the correct answer using LakeEd's permutation (2x2x4). Here I'll list them all out for you...

135
136
137
145
146
147
235
236
237
245
246
247

 

[Pikachu]

sorry, I meant 2x2x3

 

[DonW]

12 is the correct answer using LakeEd's permutation (2x2x4). Here I'll list them all out for you...

135
136
137
145
146
147
235
236
237
245
246
247

No there are many more combos Those only begin with one and two. I mean all the combos.

Don

 

[dnjan]

To make sure I understand the problem correctly -

You have three groups of numbers (or whatever).
There are two items in group 1.
There two items in group 2.
There are three items in group 3.

You are interested in the total number of combinations.

If order isn't important, and you must take one from each group, the answer is as already given - 2*2*3=12
If order is important, then for each group of three numbers, you have 3 possibilities for which number comes first, two possibilities for which number comes second, and only one number left to be last. That gives you 3*2=6 possible combinations for each set of three numbers, or 6*12=72 total combinations.

If you do not have to take a number from each group, then your possible choices will be higher yet.

 

[LakeEd]

Don,

IF you are not worried about repeating any possible combination of numbers (order not being specific), then;

2Ax2Bx3C=12
2Ax3Cx2B=12
2Bx2Ax3C=12
2Bx3Cx2A=12
3Cx2Ax2B=12
3Cx2Bx2A=12

72 possible combinations;

Where group A has 2 numbers, group B has 2 number, and group C has 3 numbers... and any of the groups may be the 1st digit number, 2nd digit number or 3rd digit number.

 

[LakeEd]

I think I understand it better the way Don (dnjan) put it.

 

[Scooterman]

arrrrg I think it is questions like this is why Boomer had to retire

 

[kendog261]

Don,

IF you are not worried about repeating any possible combination of numbers (order not being specific), then;

2Ax2Bx3C=12
2Ax3Cx2B=12
2Bx2Ax3C=12
2Bx3Cx2A=12
3Cx2Ax2B=12
3Cx2Bx2A=12

72 possible combinations;

I agree and to make it simple heres the numbers
135
136
137
145
146
147
153
154
163
164
173
174

235
236
237
245
246
247
253
254
263
264
273
274

315
316
317
325
326
327
351
352
361
362
371
372


415
416
417
425
426
427
451
452
461
462
471
472

513
514
523
524
531
532
541
542

613
614
623
624
631
632
641
642

713
714
723
724
731
732
741
742

 

[ronj]

Don, i'm only going to throw this out there once...2+2=4 !!

 

[morgan]

can i just ask what this reffers to or are u doing math problems and need everyones help lol

 

[trido]

can i just ask what this reffers to or are u doing math problems and need everyones help lol

My interpretation from my short time in this forum is that Don ( the man of many skilled talents) is simply bored and wanted to quiz everyone for his own amusement.:lol:

 

[burning2nd]

holy ----

Im gonna have to look at this after school, cause if i try and do it now ill probly have a attack of some kind

 

[DonW]

My interpretation from my short time in this forum is that Don ( the man of many skilled talents) is simply bored and wanted to quiz everyone for his own amusement.:lol:

Those numbers equate to every possible combination that my heaters and chiller combined can sequence.
Each controler is a number group, Ranco 1= one's, Ranco 2= tens, Chiller heat/cool = hundreds. I just added the heat side to the chiller so need to redo the code.

So basicly if it sees for example: ranco one on, ranco two on and chiller in cool mode the chiller is pulled offline until the the rancos cycle off.
Meaning the chiller is reading is false. All three have to match mode in order to work. If one does not match it is taken offline until the next mode cycle.

Don

 

[Elmo18]

Hi all. I did not see this thread.

This is a combinatorics/probability question.

There are 3 groups. First number can come out of any of the three "unused" groups. Second number now can only come out of the "other two unused groups", since we used 1 group already. Lastly, we have only 1 choice after taking two groups.

Hence, (3)(2)(1) = 6.

Now, we use permutation/combination, but in words, in Group 1, we can choose 1 from 2 possible. Group 2, we choose 1 from 2 possible, and from Group 3, choose 1 from 3 possible.

For example, in Group 1, "2 choose 1" is really = 2! / [(2-1)!*1!] = 2.
Similarly, in Group 2, "2 choose 1" is really = 2! / [(2-1)!*1!] = 2.
Lastly, in Group 3, "3 choose 1" is really = 3! / [(3-1)!*1!] = 3, where the "!" is factorial.

Hence, (2)(2)(3) = 12.

We multiply and get (6)(12) = 72 possible.

Best,
Ilham

 

[DonW]

Hi all. I did not see this thread.

This is a combinatorics/probability question.

There are 3 groups. First number can come out of any of the three "unused" groups. Second number now can only come out of the "other two unused groups", since we used 1 group already. Lastly, we have only 1 choice after taking two groups.

Hence, (3)(2)(1) = 6.

Now, we use permutation/combination, but in words, in Group 1, we can choose 1 from 2 possible. Group 2, we choose 1 from 2 possible, and from Group 3, choose 1 from 3 possible.

For example, in Group 1, "2 choose 1" is really = 2! / (2-1)!*1! = 2.
Similarly, in Group 2, "2 choose 1" is really = 2! / (2-1)!*1! = 2.
Lastly, in Group 3, "3 choose 1" is really = 3! / (3-1)!*1! = 3, where the "!" is factorial.

Hence, (2)(2)(3) = 12.

We multiply and get (6)(12) = 72 possible.

Best,
Ilham

Thank, the first part is what I couldnt remember. Thats why I asked here, we have plent of intelligent individuals hanging out.

Thanks again
Don