Hello !

I've found that for calculation probabelities that all boxes are filled I have to deal with Biiiiiiiiiiiig numbers. How I calculate:

For example, let's take 5 types of chocolates and 3 boxes.

All possible distinguished variants for placement chocolates is:

Box1 Box2 Box3

I I III

I II II

Let's take first placemets. All possible variants of box permutation is 3!/2! (as two boxes will have equally number of chocolates). Probabilities such one placement would be 1/(3^5). Permutation of chocolates themselves inside boxes will be 5!/3! (as one of boxes will have 3 chocolates). Well, common probabilities placement of first type is 1/(3^5)*(3!/2!)*(5!/3!)=60/243

Calculate probabilities for second type placement like for first one.

All possible variants of box permutation is 3!/2!. Probabilities such one placement would be 1/(3^5). Permutation of chocolates themselves inside boxes will be 5!/(2!*2!). Well, common probabilities placement of second type is 1/(3^5)*(3!/2!)*(5!/(2!*2!))=90/243

Common probabilities for all placement where all boxes are filled:

60/243+90/243=150/243

So probabilities at least one boxes is empty is (1-150/243).

I show easy example. But when quantity of chocolates and boxes grows, numbers become too large.

Should I use long ariphmetic ?