How to prove the amounts of dominos with x+y=n+k = x+y=n-k? I've been trying to answer this question for hours with no luck at all.. The question is the following:
Question

*

*Imagine we have domino blocks of the following shape: [ x | y ] with x, y ∈ [0..n].

*Imagine 0 ≤ k ≤ n.

Show that the amount of blocks with

*

*x + y = n - k

is equal to the amount of blocks with

*

*x + y = n + k

The question also says:
In both cases, you will find (n - k + 1)/2, but you need to prove this.
Thanks a lot for the help!
 A: We have that $x+y=n-k$ .Also we have $x>=0$ and $y>=0$ . Then we can get that $n-k>=x>=0$ . For each value of x we have one domino so we write $(n-k+1)$ . If $2|(n-k+1)$ than each domino was counted two times so answer is $(n-k+1)/2$ . If $2|(n-k)$ than each domino counted two times , exept domino $((n-k)/2;(n-k)/2)$ so answer is $(n-k)/2+1$ . When $x+y=n+k$  we can write that $n>=x>=k$ and we shall get the same answer . 
A: For each $x$ from $0$ to $n-k$ let $y=n-k-x$. This makes $n-k+1$ blocks with $x+y=n-k$. But if $n-k+1$ is even, we must count only a half of them, because the block $[x|y]$ is the same as $[y|x]$. And if $x-k+1$ is odd, there is a block that has no pair, namely $[(x-k)/2, (x-k)/2]$.
For each block $[x|y]$ with $x+y=n-k$ we have exactly one block $[n-x|n-y]$, and $n-x+n-y=2n-(n-k)=n+k$, so the number of blocks with $x+y=n+k$ is the same as before.
A: Consider a domino $(x, y)$ such that $x+y = n-k$. Now, consider another domino $(n-x, n-y)$. We have $n-x + n-y = 2n - (x+y) = n+k$. Now, we can map dominoes $(x, y)$ that sums to $n-k$ to dominoes $(n-x, n-y)$. This mapping is one-to-one and onto (incidentally the inverse of this mapping is itself), hence it is a one to one correspondence. Thus, the number of dominoes that sums $n-k$ and $n+k$ must be the same.
