MAT Q5 1996 Solution Verification (Combinatorics) I have been doing some MAT past papers and I have got to papers where there is no official mark scheme. I checked on the student room but there seem to be no answers for the very old mat past papers. Here is the question I want to answer:

My solution:
a): There could be two values for each of the $4\times4=16$ places. Thus the total number of different arrangements is $2^{16}$
b): Picture the $4$ rows and let us consider the cases: " one $\times$ in each row", two $\times$ in each row, "three $\times$ in each row" and "four $\times$ in each row".
In the first case, we have $(^4C_1)^4$ different arrangements.
In the second case, we have $(^4C_2)^4$ different arrangements.
In the third case, we have $(^4C_3)^4$ different arrangements.
In the fourth case, we have $(^4C_4)^4$ different arrangements.
And thus the total number of different arrangements is $=1809$ (adding all of the above).
c): Consider the cases
$$\begin{pmatrix}
  \times & \text{either} & \text{either} & 0 &\\ 
   \text{either}& \times & 0 & \text{either} & \\
   \text{either} & 0 & \times & \text{either} &\\ 
   0& \text{either} & \text{either} & \times & \\
\end{pmatrix}
\text{and}
\begin{pmatrix}
  0 & \text{either} & \text{either} & \times &\\ 
   \text{either}& 0 & \times & \text{either} & \\
   \text{either} & \times & 0 & \text{either} &\\ 
   \times& \text{either} & \text{either} & 0 & \\
\end{pmatrix}
$$
In each of the cases, we have $8$ available positions giving $2^8$ possible arrangements. We need only now consider the case when
$$\begin{pmatrix}
  \times & \text{either} & \text{either} & \times &\\ 
   \text{either}& \times & \times & \text{either} & \\
   \text{either} & \times &  \times &\text{either} &\\ 
   \times& \text{either} & \text{either} & \times & \\
\end{pmatrix}$$
in which case we have yet again $2^8$ different arrangements. Thus the total number of different arrangements is $3\times 2^8=768$
Is this correct?
EDIT: I have realised I have interpreted the question way more generally than I should have. I thought that for each position I have a nought and across available. Anyways, I would like to ask if my generalization is correct.
 A: I’m afraid that all of these are incorrect.
$2^{16}$ is the number of different ways to put either a nought or a cross in each of the $16$ positions without any restriction on the number of crosses. Here, however, we must have exactly $4$ crosses; there are $\binom{16}4$ ways to choose $4$ of the $16$ positions to get the $4$ crosses, and once we’ve done that, the other $12$ positions must be filled with noughts, so the correct answer to (a) is $\binom{16}4=1820$.
In (b) we must choose one position in each row. In any one row there are $4$ possible choices, so there are altogether $4^4=256$ ways to choose one position in each row to get the cross for that row.
In (c) there are $4$ ways to place a cross in the first row. Once that’s been done, there are only $3$ possible places for the cross in the second row, since it cannot lie in the same column as the cross in the first row. Similarly, once those two crosses have been placed, there are just $2$ possible positions for the cross in the third row, and after that there is just one possible position for the cross in the fourth row. Thus, there are $4\cdot3\cdot2\cdot1=4!=24$ possible arrangements of this type.
A: Put the rows in line: you get a line of $16$ places divided in $4$ sectors, then
a) $\binom{16}{4}$ : you can put the four $\times$ in every place;
b) $4^4$ : four choices for each sector;
c) $4!$ : you can chose any permutation of $(1,2,3,4)$ and assign as place in first, 2nd , .. sector
A: No. 
Hints:
For the first one you are allowing more than $4$ crosses. So you have to choose where are the crosses out of $16$ possibilities.
For the second one, I am really confused in how you are arguing. Where does the power of the combinatorial comes from? Notice that you have $4$ choices per row. Multiplication principle gives you that...
For the third one you can not have two crosses in the same column. So here are $4$ choices for the first row but $3$ for the second one...So multiplication principle again gives you...

Edit:
For the new problem, the first one is fine. The second one is not right. Notice that per row there are $2^4-1$ possibilities (this follows from your logic in problem 1). So you have $(2^4-1)^4.$
The third one is tricky cause you will have to do inclusion-exclusion. Try to do something like $$2^{16}-(4+4)*2^{16-4}+2\cdot \binom{4}{2}\cdot 2^{16-8}+4\cdot 4\cdot 2^{16-8+1}\cdots,$$
Essentially you are doing all of them and excluding when one row or one column has no crosses.
