Just a normal arrangement problem!

Since, none of the box can't be kept empty, there are only two possibilties in which we can distribute the balls in 3 boxes. It is (3,1,1) or (2,2,1). And as the boxes and balls are of different size and colour respectively, we can arrange them among themselves in 3 ways. So total number of ways is, $( \dbinom{5}{3} \dbinom{3}{1} \dbinom{1}{1} + \dbinom{5}{2} \dbinom{3}{2} \dbinom{1}{1} ) 3$ $= \boxed{150}$ Hence, N = 150 so answer is, $\dfrac{150}{30} + 4! = \boxed{29}$

Use the Stirling numbers of the second kind sequence where S(k,n) n! where k is number of balls and n number of boxes. In this case, S(5,3)=25 3!=150 ways. Since the answer is N/30 + 4!, answer is 29

Let's give boxes a name, A B and C.

First case: (3,1,1)

• Choose 3 balls from 5 balls for box A, $\large\binom{5}{3}$ .
• Choose 1 ball from 2 balls for box B, $\large\binom{2}{1}$
• Choose 1 ball from 1 ball for box C, $\large\binom{1}{1}$
• You can choose box A or B or C to have 3 balls and the other have 1 ball, $\large\binom{3}{2}$

This case will have $\large\binom{5}{3}\large\binom{2}{1}\large\binom{1}{1}\large\binom{3}{2} = 60$ ways.

Second case: (2,2,1)

• Choose 2 balls from 5 balls for box A, $\large\binom{5}{2}$ .
• Choose 2 balls from 3 balls for box B, $\large\binom{3}{2}$
• Choose 1 ball from 1 ball for box C, $\large\binom{1}{1}$
• You can choose box A or B or C to have 1 ball and the other have 2 balls, $\large\binom{3}{2}$

This case will have $\large\binom{5}{2}\large\binom{3}{2}\large\binom{1}{1}\large\binom{3}{2} = 90$ ways.

Therefore, there're total of $60+90 = 150$ ways to do that.

Answer: $\boxed{29}$ . :3