true\false questions on number of solutions in combinatorics I was preparing for the exam, and then I saw a few questions I did not understand regarding number of solutions in combinatorics. I would be glad if you could take a look and correct me if I'm wrong:
(questions are true\false):
1) The number of natural solutions of $x_1+x_2+x_3+x_4=14$ is equal to the number of positive whole number (integer) solutions of: $x_1+x_2+x_3+x_4=10$
2) The number of the positive even natural number solutions of $x_1+x_2+x_3+x_4=14$ is equal to the number of the positive odd natural number solutions of $x_1+x_2+x_3+x_4=14$
3) The number of integer solutions that are either $1$ or $-1$ to $x_1+x_2+x_3+x_4 \geq 0$ is 10.
4) If $|a|=3^3$ and $|b|=3$, then the number of functions from $a$ to $b$ is equal to the number of functions from $b$ to $a$.
What I did:
1)  False.  The number of the natural solutions of $x_1+x_2+x_3+x_4=14$ is $\binom {14+4-1}{3}$ and the number number of positive integer solutions of $x_1+x_2+x_3+x_4=10$ is $\binom {10+4-1}{3}$. It's false because the results are not the same.
2)  False. The number of positive odd natural solutions of $x_1+x_2+x_3+x_4=14$ can be obtained if we divided the result by $2$ ($y_i=x_i/2$), i.e: $y_1+y_2+y_3+y_4=7$, so $\binom {7+4-1}{3}$. and for the even, we need that $x_i \geq 1$, so $y_i=\frac{x_i-1}{2}$, so $\binom {5+4-1}{3}$.
3)  I don't know how to solve it, but I'll try.  If we said the solution is only $1$, then there were no limits, and the result would be: $4!=24$. However, since it's more larger than $0$, we must have more $1$s than $-1$s, so we have to subtract the cases were we have more $-1$ than $1$, i.e: if at least $3$ variables are $-1$, so we subtract $2$ possibilities (either $3$ are $-1$ or $4$ are $-1$), and we get that the number of solutions is $24-2=22$ (If it was $3!-2$, it would make sense. I think I did it wrong).
4) False. If $|a|=3^3=27$ and $|b|=3$, then the number of functions from $a$ to $b$ is not the same as the number of functions from $b$ to $a$.
Please help me and correct me if I made any mistakes. Did my best to elaborate and explain what I've done. Thank you in advance!
 A: 1) FALSE
$$
\eqalign{
  & \quad \quad \left( {x_{\,1}  + 1} \right) + \left( {x_{\,2}  + 1} \right) + \left( {x_{\,3}  + 1} \right) + \left( {x_{\,4}  + 1} \right)\quad \left| {\;0 \le x_{\,k}  \in Z} \right.\quad  \Leftrightarrow   \cr 
  &  \Leftrightarrow \quad x_{\,1}  + x_{\,2}  + x_{\,3}  + x_{\,4} \quad \left| {\;1 \le x_{\,k}  \in Z} \right. \cr} 
$$
It would be true the reverse (exchanging 10 and 14).
2) FALSE
$$
\eqalign{
  & {\rm N}{\rm .}\,{\rm sol}{\rm .}\,{\rm of}\;\left\{ {2x_{\,1}  + 2x_{\,2}  + 2x_{\,3}  + 2x_{\,4}  = 14\quad  \Rightarrow \quad x_{\,1}  + x_{\,2}  + x_{\,3}  + x_{\,4}  = 7\quad \left| {\;1 \le x_{\,k}  \in Z} \right.} \right\} <   \cr 
  &  < {\rm N}{\rm .}\,{\rm sol}{\rm .}\,{\rm of}\;\left\{ {2x_{\,1}  - 1 + 2x_{\,2}  - 1 + 2x_{\,3}  - 1 + 2x_{\,4}  - 1 = 14\quad  \Rightarrow \quad x_{\,1}  + x_{\,2}  + x_{\,3}  + x_{\,4}  = 9\quad \left| {\;\;1 \le x_{\,k}  \in Z} \right.} \right\} \cr} 
$$
3) FALSE
The number of $-1$'s can be $0$ or $1$ or $2$, corresponding to ${ 4 \choose 0}+{ 4 \choose 1}+{ 4 \choose 2}=11$ ways to place them.
4)  FALSE
Understanding $|a|$ as the number of elements of the set $a$, then the functions $f:a \to b$ are $|b|^{|a|}$ and those $f:b \to a$ are $|a|^{|b|}$   
Thanks to N.F. Taussig for pointing out the correct interpretation of  question (4), and other misunderstandings. 
