bijection of the sum of 3 variables

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I was doing an exercice and there is a sentence in the correction i didn't quite understand. I had three balanced dices and I had to find the probability that the sum gave an even result. When I check my answer they just give a weird explanation : let $(i,j,k)$ be result of my 3 dices in $\{1,2,3,4,5,6\}$ and an application $\omega \mapsto (7-i,7-j,7-k)$ is said to be the bijection of the $(i,j,k)$ giving an even sum over those over an uneven onesum.

I know what is a bijection but I just don't get it why it's a bijection in between : {sum of dice is even} and {sum of dice is uneven}

I'd just like to understand because it doesn't make sense to me.

Thank you for your help

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It's one-to-one, each triple $(7-i,7-j,7-k)$ corresponds to a single input $(i,j,k)$. There's no way to have $(7-i_1,7-j_1,7-k_1)=(7-i_2,7-j_2,7-k_2)$ unles $(i_1,j_1,k_1) = (i_2,j_2,k_2)$. Furthermore, it is surjective (onto) since we have when $(7-1,7-1,7-1,)=(6,6,6)$ all the way to $(7-6,7-6,7-6)=(1,1,1)$, i.e. all the possible triples $(i,j,k)$ are accounted for. Does that make sense?

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If $i+j+k$ is even $(7-i)+(7-j)+(7-k)$ is odd.