How many solutions are there for the inequation:
$$|X_{1}|+|X_{2}|+\cdots+|X_{k}|\le n$$ For every $i$, $X_i\in\mathbb{Z}$
My attempt:
I know this problem $X_{1}+X_{2}+\cdots+X_{k}\le n$ $(X_i\ge0)$ is equal to $X_{1}+X_{2}+\cdots+X_{k+1}=n$. Now we have to decide for every $X_i$ whether it is $X_i$,$(-X_i)$ or zero. Therefore:
$PP_{3}^{k+1}\cdot CC_{k+1}^{n}=3^{k+1}\cdot\binom{n+k}{n}$
But It is not the correct answer.
Where is my mistake?
Your answer implies that you first choose a solution to $X_1+\cdots+X_{k+1}=n$ with $X_i\geq 0$, and then make one of three choices (positive, negative, or zero) for each $X_i$.
The problem here is that you can't make any choice for any $X_i$. If $X_i=0$ in your solution, then it just is zero and no further choice can be made. If $X_i>0$, then you can decide whether to make it positive or negative (two choices), but not zero.