no rows or columns contain same letters, is 
The number of ways in which we can fill $a,a$ and $b,b$ in $4\times 4$ grid such that each cell contain  at most one letter and no rows or columns contain same letters, is

what i try
Total number of ways $\displaystyle =\binom{16}{4}\cdot \frac{4!}{2!\cdot2!}=10920$
(above selceting 4 cells out of $16$ and arrange $2a,2b$ in these cells)
$\bullet\; $ both letters are together $\displaystyle \binom{8}{1}\cdot \binom{4}{2}\cdot 1$
( above selecting one row our colum from total $8$ and each row or column contain $4$ cells we can select $2$ out of $4$  )
Did not know how can i  solve. 
 help me please 
 A: let's first find ways of arranging 'a' first  . Let's say you first 'a' in any of the cells , now for the 2nd 'a' only 9 cells are left(because 3 cells each of first 'a'$^s$ column and row cannot be taken, i will refer to it as destroying cells, so it destroyed 6 cells). But in counting this way every case is repeated 2  times . $\frac{16\times9}{2}=72$

Now if you decide to put both  b   in such a way it does not share any row or column with a . (2nd 'a' destroys 4 more cells as 2 of the cells been destroyed are shared by them. 2 are occupied by a). So we are left with 4 cells . Now putting first b in any cells , destroy 2 more cells more . Leaving only one cell for next b . But counting this way every case is repeated 2 times. So we get 2 ways from here.

one b shares one column or 1 row with (not 2) with one a and other b do not share any row or column with a. There 8 cells which shares one row or column with a . Putting b in one of those , destroy 2 cells for other b (originally 4) . No case is counted twice. So from here we get $8\times2$ cases=16

one b shares 2 column and row and other b shares no row or column(it's not possible for b to share one column and one row with same a  , neither it's possible for b to share both rows with a . ) There are 2 cells such that shares one row and one column with both a. putting first b in those does not destroy any of the 4 cells . Hence from we get 8 cases.

one b shares 2 column and row and one b shares one row or column. Putting first (sharing 2 things ) destroy 4 cells . hence from 8 cases

both b shares one row or column. Putting first b destroy 2 cells giving 5 options for 2nd = $\frac{8\times5}{2}$ (each case counted 2 times)= 20 ways.

both b shares 2 row and column . 1 way only.

total ways of selecting places for b= 2+16+8+8+20+1=55ways

Final answer $55\times72=3960$

alternatively , this is really better.
there would have been 72 cases if the a's were not present. we will subract the cases in which one b occupies any one a position. Let's put first b in place of first a giving us 9 cases . And 9 cases if 2nd a position occupied by first b. But the case with both b occupying both a position is counted twice giving us  17. Ways of arranging b once both a have been arranged = 72-17=55.
