Expert: Paul Klarreich - 11/16/2009

Question
Consider the following variables:
                      
                      Z
              A1      B2       C3
            A4  D1   B5  E2   C6  F3
            G4  D7   H5  E8   I6  F9

How many groups can be made with the variables from the set if two
variables with the same number or same letter cannot be in the same set and Z cannot be with A1, B2, or C3 (include the empty set)

Now I think I got the answer. Can you just check my work to make sure its right?

Okay. First I split the variables up like so:
             GROUP 1   GROUP 2  GROUP 3   
              A1     |   B2    |   C3
            A4 D1    |  B5 E2  |  C6 F3
            G4 D7    |  H5 E8  |  I6 F9

I ignored Z for the moment. My motivation for splitting the variables up like that was because any variable from GROUP 1 wont affect GROUP 2 or GROUP 3, so we can find the total number of groups in just one group and cube it since the same number of groups will be in every other group.

If we do GROUP 1, simple listing gives the groups

{empty set}
A1
D1
G4
D7
A4
A1 G4
A1 D7
A4 D1
A4 D7
D1 G4
G4 D7
A1 G4 D7

For a total of 13 (THOUGH I PROBABLY MISCOUNTED-CAN YOU VERIFY THIS?).
By symmetry, there will then be 13 groups in GROUP 2 and 13 groups in GROUP3, for a total of 2197.

But that is not including Z in any groups. Z cannot be with A1, B2, or C3, and we see that there are 4 groups in GROUP 1 that include A1, so by symmetry there will be 4 groups in GROUP 2 that include B2, and four groups in GROUP 3 that include C3. Hence, there are only (13-4)^3=729 groups which can have the Z added to it.

So the total number of groups is 13^3+9^3=2926



Can you PLEASE tell me if there is a mistake in there somewhere?

Answer
Questioner: Ryan
Country: United States
Category: Advanced Math
Private: No
Subject: Easy Counting Problem?
Question: Consider the following variables:
                     
                     Z
             A1      B2       C3
           A4  D1   B5  E2   C6  F3
           G4  D7   H5  E8   I6  F9

How many groups

>> (do you mean 'subsets'?)
can be made with the variables from the set if two
variables with the same number or same letter cannot be in the same set and Z cannot be with A1, B2, or C3 (include the empty set)

Now I think I got the answer. Can you just check my work to make sure its right?

Okay. First I split the variables up like so:
            GROUP 1   GROUP 2  GROUP 3   
             A1     |   B2    |   C3
           A4 D1    |  B5 E2  |  C6 F3
           G4 D7    |  H5 E8  |  I6 F9

I ignored Z for the moment. My motivation for splitting the variables up like that was because any variable from GROUP 1 wont affect GROUP 2 or GROUP 3,

>> What do you mean:  "won't affect"?  Do you mean to say that "Any member of Gr1 can be combined with any member of Gr2 or Gr3?"

so we can find the total number of groups in just one group and cube it since the same number of groups will be in every other group.

If we do GROUP 1, simple listing gives the groups

{empty set}
A1
D1
G4
D7
A4
A1 G4
A1 D7
A4 D1
A4 D7
D1 G4
G4 D7
A1 G4 D7

For a total of 13 (THOUGH I PROBABLY MISCOUNTED-CAN YOU VERIFY THIS?).

>> yes, it looks right.

By symmetry, there will then be 13 groups in GROUP 2 and 13 groups in GROUP3, for a total of 2197.

But that is not including Z in any groups. Z cannot be with A1, B2, or C3, and we see that there are 4 groups in GROUP 1 that include A1, so by symmetry there will be 4 groups in GROUP 2 that include B2, and four groups in GROUP 3 that include C3. Hence, there are only (13-4)^3=729 groups which can have the Z added to it.

So the total number of groups is 13^3+9^3=2926

>> Yes.  I don't exactly like your vocabulary, but your reasoning looks good.