How is this code's answer 10?

Question

#define prod(i,j)i*j
int main(){
int x=3,y=4;
printf("%d",prod( x+2, y-1));
return 0;
}

Answer 1

I think you copied the code wrong. Instead of
printf( "%d", ( x + 2, y - 1 ) );
I would guess the original line was
printf( "%d", prod( x + 2, y - 1 ) );
otherwise the result would be three, not ten.

Anyway, as to how the answer is ten, the explanation is that #define is a literal text replacement. Whenever the macro is encountered during precompilation, it is replaced by exactly the macro content.
Therefore, the print statement becomes
printf( "%d", x + 2 * y - 1 );
and from there, it is simply operator precedence that needs to be taken into account, i.e. multiplication before addition.

On a sidenote, please don't tag languages that are irrelevant to your question.

Answer 2

Thank you❤
I edited it..