The wording of this question is quite subtle. It ask for us to count the number of ways that the balls can be put in the box such that any two boxes have more balls than the third.

I assume the question refers to using all the balls i.e. no ball can be left outside a box.

Hence in order for this requirement to be fulfilled no single box can have more than half the balls i.e. since there are

balls in total the maximum number of balls we can place in any single box must be

. It should be easy to see that this guarantees the requirement is fulfilled.

Now let's call the three distinct boxes

,

and

and the respective number of balls placed in each

,

and

.

If we start by placing a ball in

then

and

must have

balls each i.e. using the notation

to signify the different configurations we have that when

the configuration must be

and this is the only possible configuration for

since exchanging a ball from

to

would result in

and vice versa.

Now it should be easy to see that when we have

(

) one configuration we have is

and we can make

exchanges from

to

so in total (including the above configuration) we have

configurations of the balls when

.

Hence the total number of configurations,

, is given by

So using the well known result for this arithmetic progression we have that

.