I'm not quite sure abouth this one.
1st row can be formed in 3x3x3 ways and to make their position fixed, they can be arranged in 3 ways.So total is 81.
Let's say the 1st row looks like this now
A B C
X X X
X X X

Now 1st column can be chosen in 2x2 (from 2B & 2C)=4 ways and themselves cna be arranged in 2 ways. So total is 81*8.

A B C
B X X
C X X

2nd column can be chosen in 1X2 ( 1C & 2A) ways and among themselve 2 ways. So total is 81*8*4

A B C
B C X
C A X

3rd column can be chosen in 1x1 (1A & 1B) ways and among themselve in 2 ways,

A B C
B C A
C A B
Total is 81*8*4*2=5184 (Seems to big!)

Way too big indeed

Are you sure there are 81 ways to form the first row?

[email protected] wrote:I'm not quite sure abouth this one.
1st row can be formed in 3x3x3 ways and to make their position fixed, they can be arranged in 3 ways.So total is 81.
Let's say the 1st row looks like this now
A B C
X X X
X X X

Now 1st column can be chosen in 2x2 (from 2B & 2C)=4 ways and themselves cna be arranged in 2 ways. So total is 81*8.

A B C
B X X
C X X

2nd column can be chosen in 1X2 ( 1C & 2A) ways and among themselve 2 ways. So total is 81*8*4

A B C
B C X
C A X

3rd column can be chosen in 1x1 (1A & 1B) ways and among themselve in 2 ways,

A B C
B C A
C A B
Total is 81*8*4*2=5184 (Seems to big!)

I'm bit confused in this problem.
If we consider all the A,B & C alphabets a a unit then 1st row can be formed in:
3C1*3+2C1*3+1C1*3 ways..
Am i wrong again?

In how many different ways can 3 As, 3Bs and 3Cs be arranged in a 3x3 matrix so that no row or column has more than one of each of the alphabets?

1) Choose first A to place, you have 9 places to choose: 9 ways
2) choose second A to place , you have 4 places to choose : 4 ways
3) choose 3rd A to place , you have only 1 place to choose : 1 way

Thus 9 X 4 X 1 ways for all A's

similarly for B and C

6 X 3 X 1 and 3 X 2 X 1 ways

Thus total ways: 9 X 4 X 6 X 3 X 3 X 2
also All As, Bs and Cs can be interchanged in their position thus we need to divide this by 3! three times.
that gives us 18 as answer!

correct me if I'm wrong!

santhoshsram wrote:In how many different ways can 3 As, 3Bs and 3Cs be arranged in a 3x3 matrix so that no row or column has more than one of each of the alphabets?

it will be difficult to answer without choices.
it sometimes allows us to ballpark few choices.
IMO [spoiler]36?[/spoiler]

user123321

santhoshsram wrote:In how many different ways can 3 As, 3Bs and 3Cs be arranged in a 3x3 matrix so that no row or column has more than one of each of the alphabets?

Top row:
A, B and C must each appear once in the top row.
Number of ways to arrange A, B and C = 3! = 6.

Second row:
Number of options for A = 2. (Either of the 2 columns not occupied by A in the top row.)
Once A is placed in the second row, there is only ONE acceptable arrangement for B and C in the second row.

To illustrate:
ABC
XAX

implies

ABC
CAB

Thus, once A has been placed, the number of options for B and C = 1.
To combine our options for the 2nd row, we multiply:
2*1 = 2.

Bottom row:
Once A, B and C are placed in the top two rows, there is only ONE acceptable arrangement for the bottom row.

To illustrate:
ABC
CAB

implies

ABC
CAB
BCA

Thus, the number of options for the bottom row = 1.

Total number of possible arrangements:
To combine our options for each row, we multiply:
6*2*1 = 12.

An alternate approach would be to write out all the possible options:

ABC...ABC...ACB...ACB
CAB...BCA...BAC...CBA
BCA...CAB...CBA...BAC

BAC...BAC...CAB...CAB
ACB...CBA...ABC...BCA
CBA...ACB...BCA...ABC

BCA...BCA...CBA...CBA
ABC...CAB...ACB...BAC
CAB...ABC...BAC...ACB

Crystal clear explanation from Mitch, as always. Thanks Mitch.

CLUE: For each COLUMN and ROW the other two COLUMNS or ROWS can happen in 2 ways.

So for Column: 2+2+2=6 ways
and for Rows: = 2+2+2=6

Total=6+6=12 ways