To solve the problem, we need to account for the condition that Hugh (H) refuses to stand anywhere in between Louise (L) and Richard (R). Let's denote the five children as H, L, R, and two other children (let's say A and B).

First, we calculate the total number of arrangements of the five children without any restrictions. The number of ways to arrange 5 children is given by:

\[
5! = 120
\]

Next, we need to determine what arrangements we need to exclude—those where Hugh is standing between Louise and Richard.

To find out the specific arrangements where Hugh is between L and R, we can consider L and R as creating a single block (L, H, R or R, H, L). This means we can treat (L, H, R) as a single unit or block, which we can denote as (LHR). Thus, our blocks now become:

- (LHR) or (RHL)
- A
- B

Now we have 4 units to arrange: (LHR) or (RHL), A, and B. The number of ways to arrange these 4 units is:

\[
4! = 24
\]

However, there are two configurations for the block itself (LHR or RHL). So, we need to multiply by 2:

\[
4! \times 2 = 24 \times 2 = 48
\]

Therefore, the number of arrangements where Hugh is between Louise and Richard is 48.

Finally, we subtract this value from the total arrangements to find the number of arrangements that keep Hugh happy (i.e., not between Louise and Richard):

\[
120 - 48 = 72
\]

Thus, the total number of ways to arrange the children in a line while keeping Hugh happy is:

\[
\boxed{72}
\]