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
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