The combination C(y,y) is equal to 1.
The formula for the number of combinations of n items taken r at a time is:
C(n,r) = n! / (r!(n-r)!)
In this case, we have:
C(y,y) = y! / (y!(y-y)!)
Since y - y = 0, we can simplify the expression to:
C(y,y) = y! / (y! * 0!)
And since 0! = 1, we can further simplify to:
C(y,y) = y! / y!
The y! term in the numerator and denominator cancel out, leaving us with:
C(y,y) = 1
Therefore, the combination C(y,y) is equal to 1.