Balls and Boxes gone crazy. Part-I

Total ways of placing $5$ balls in $3$ boxes $= 3^5$

Total ways in which $1$ box is empty $= 3 \times 2^5$

Total ways in which $2$ boxes are empty $= 3$

Using principle of inclusion and exclusion:

Total ways of placing balls so that no box remains empty $= 3^5 - 3 \times 2^5 + 3 = 150$

In other words, we have to divide 5 distinct balls to 3 distinct boxes such that no box is remained empty. Possible configuration for boxes are: permutations of (2,2,1) and (3,1,1) .

So, total number of ways of doing (2,2,1) divisions is (5!/2! 2! 1!)/2!. Now, this is to be arranged in 3 boxes, so total number of ways 15 * 3! = 90

Similarly, total number of ways of doing (3,1,1) divisions is (5!/3! 1! 1!)/2!. Now, this is to be arranged in 3 boxes, so total number of ways 10 * 3! = 60

So, by Fundamental principle of counting, total number of ways = 90+60 = 150

We want to compute the number of surjections (since no box ought to remain empty) from the set $⟦1,5⟧$ of numbered balls to the set $⟦1,3⟧$ of numbered boxes.

And it is well known that the number of surjections from $⟦1,n⟧$ to $⟦1,m⟧$ (with $n≥m$ ) is

$\underbrace{m^n}_{\text{cardinal of } ⟦1,m⟧^{⟦1,n⟧} } - \Bigg( \underbrace{ \sum\limits_{k=1}^m (-1)^{k-1} \binom{m}{k} (m-k)^n }_{ \text{ cardinal of } \bigcup\limits_{i=1}^m E_i \, \, , \text{ where } E_i \text{ is the set of functions of } ⟦1,m⟧^{⟦1,n⟧} \text{ that don't take the value } i } \Bigg)$

Hence the wanted number is :

$3^5 - 3 \times 2^5 + 3 \times 1 = 150$