number of multiplication steps required to solve Ax = b

Question

If we can factorize $A$ in $LU$, We can solve $Ax = b$ in 2 steps:

1. Solve $Lc = b$ for c
2. Solve $Ux = c$ for x

As per the Linear algebra book by Gilbert Strang, each step takes $n^2/2$ number of multiplication-subtraction. I am able to find that for 2nd step, but not for the first step. For $Lc = b$, I can see that there are $(n^2 - n)$ no. of multiplication-subtractions.

Can anyone please help me finding how we have $(n^2/2)$ calculations ?

Thanks.