Find the probability that the sum of the numbers in at least one horizontal row is greater than $21$. Nine numbers $1,2,3, \ldots, 9$ are arranged in a rectangular array of matrix of order $3$ so that each number occurs exactly once.  Find the probability that the sum of the numbers in atleast one horizontal row is greater than $21$.
I counted the probability as $\frac{18+18+18}{9!}$ but the correct answer is $\frac{1}{7}$.  Can you please help me with correct approach to solve this?
 A: Hints:


*

*Show that a row with sum greater than $21$ must contain $9$

*Count how many possibilities there are for the other two values in the row containing $9$

*Count how many of those have a sum greater than $21$

*Divide

A: 1) There are only four combinations of three numbers so that the sum is greater than 21.These combinations are (9 8 7);(9 8 6);(9 8 5);(9 7 6).
2) Now,each combinations can be permuted(arranged) in 3! ways.Hence the total number of possible arrangements will be (4*3!).
3) Now,these different arrangements can be placed at three different rows of the matrix(R1,R2,R3).And hence the total number of possibilities of placing the different combinations at different places will be equal to (3*4*3!).
4) Now,the row containing 3 numbers which is going to produce the total of 12 is fixed and we can arrange the other six elements of the matrix in 6! ways.Hence the total number of different possible matrices will be equal to (3*4*3!*6!) and this our favorable outcome.
5) Our sample space will have 9! elements,no need to explain this.
Hence probability= (3*4*3!*6!)/9! = 1/7
