How many $3\times 3$ arrays with digits from $1$ to $9$ with increasing order are there? The entries in a $3 \times 3$ array include all the digits from $1$ through $9$, arranged so that the entries in every row and column are in increasing order. How many such arrays are there?
This is a question on combinatorics.
I tried using tableaus and using hook numbers but couldn't understand after that please tell how to solve this.
It would be easier for me if solved by using normal combinatorics. But no restrictions. It's your choice
 A: Using the notation $(A,B,C)$ to describe the number $C$ being located in the $A$ row and $B$ column.  Due to symmetry, the transpose (reflection across the main diagonal) of any solution is a different solution, in other words, if we have a solution:
$$\{(A_1,B_1,1), (A_2,B_2,2), (A_3,B_3,3), (A_4,B_4,4), (A_5,B_5,5), (A_6,B_6,6), (A_7,B_7,7), (A_8,B_8,8), (A_9,B_9,9)\}$$
then we also have a solution:
$$\{(B_1,A_1,1), (B_2,A_2,2), (B_3,A_3,3), (B_4,A_4,4), (B_5,A_5,5), (B_6,A_6,6), (B_7,A_7,7), (B_8,A_8,8), (B_9,A_9,9)\}$$
Because every row and column has to be in increasing order, we know that our solution has to include $(1,1,1)$ and $(3,3,9)$.
We have two choices for where to put the number $8$. Due to symmetry, we will consider only the solutions with $(3,2,8)$, and will just need to double the number of solutions.
We now have two choices for where to put $7$:
Case 1: $(3,1,7)$
The number $6$ is locked in as $(2,3,6)$.  The number $5$ can be either in $(2,2,5)$ or $(1,3,5)$.  If $(2,2,5)$, then the numbers $2,3,4$ have to be in the three remaining spots; as soon as we pick which one is in $(2,1,X)$, then the rest are locked in place, giving three solutions with $(3,1,7)$ and $(2,2,5)$. If $(1,3,5)$, then we must have $(2,2,4)$, and have only either $(1,2,2)$ and $(2,1,3)$ or $(1,2,3)$ and $(2,1,2)$ for another two solutions.
Case 2: $(2,3,7)$
The numbers $5$ and $6$ must be in two of the three spots of the main antidiagonal (the top right, the middle square, and the bottom left).  The are therefore $3!=6$ ways of assigning them.  In the two cases where neither one is in the middle space, the number $4$ must be in the middle space, and there are two possible arrangements for the numbers $2$ and $3$.  In the each of the other four cases, there are two cases where the number $4$ is in the remaining space on the main antidiagonal and one where it isn't.  This results n a total of 16 arrangements if $(2,3,7)$.
Therefore, the total number of arrangements is $2(3+2+2\cdot2+4\cdot3)=42$
A: The $1$ and the $9$ must clearly go in the upper left hand and lower right hand corners, respectively. It's easy to see that the $5$ cannot be adjacent to either the $1$ or the $9$, hence it must go in one of the three spots on the "anti" diagonal. Inventing a bit of notation, we can write the number of possibilities as
$$\#\pmatrix{1&*&\\*&5&-\\&-&9}+2\times\#\pmatrix{1&*&5\\*&&-\\&-&9}$$
where the "$\#$" of a $3\times3$ array denotes the number of solutions with $1$, $5$, and $9$ in assigned spots, with each $*$ understood to be a number between $1$ and $5$ and each $-$ a number between $5$ and $9$. The "$2\times\,$" is for the symmetry that would have the $5$ in the lower left hand corner. By the same symmetry, we have
$$\#\pmatrix{1&*&\\*&5&-\\&-&9}=2\times\#\pmatrix{1&*&*\\*&5&-\\-&-&9}$$
and it's now easy to see that the three $*$'s can be filled with the numbers $2$, $3$, and $4$ in just $3$ different ways, and likewise for the three $-$'s with the numbers $6$, $7$, and $8$, so that
$$\#\pmatrix{1&*&\\*&5&-\\&-&9}=2\times3\times3$$
A somewhat different symmetry argument tells us
$$\#\pmatrix{1&*&5\\*&&-\\&-&9}=2\times\#\pmatrix{1&*&5\\*&*&-\\-&-&9}$$
and in this case now the $4$ has only one spot it can go into:
$$\#\pmatrix{1&*&5\\*&*&-\\-&-&9}=\#\pmatrix{1&*&5\\*&4&-\\-&-&9}=2\times3$$
Putting everything together, the total number of arrangements is
$$(2\times3\times3)+(2\times2\times\times2\times3)=18+24=42$$
Remark (added later): For clarity and precision, the "somewhat different" symmetry that tells us
$$\#\pmatrix{1&*&5\\*&*&-\\-&-&9}=\#\pmatrix{1&*&5\\*&-&-\\*&-&9}$$
is a reflection across the "anti" diagonal followed (or preceded) by the numerical replacement $k\to10-k$ for each $k\in\{1,2,3,4,5,6,7,8,9\}$.
