Question

Eight identical balls are to be placed in three different boxes. If each box can hold any number of balls, then in how many ways can the balls be placed in the boxes so that no box remains empty?

Answer 1

There are 21 different ways to place eight identical balls into three different boxes such that no box remains empty, using the stars and bars combinatorial method.

Step-by-step explanation:

To determine the number of ways to place eight identical balls into three different boxes such that no box remains empty, we need to use combinatorics and account for the constraint that every box must contain at least one ball. Since the balls are identical, the order in which they are placed doesn't matter, simplifying the calculation. A classic approach to solving such problems is to use stars and bars combinatorics.

• If we denote a ball by a star (*) and a separation between boxes by a bar (|), we are essentially looking for all the possible ways to arrange seven stars and two bars.
• This arrangement forms a simple sequence of nine places where the stars can be.
• Since we do not want any box to be empty, we first place one star in each box.
• We now have five stars left to distribute freely among the three boxes.
• The number of ways to arrange these five stars and two bars that separate the three regions is the number of combinations we can choose five positions from seven, which is given by the binomial coefficient C(7,5).

21 ways (C(7,5) = 7! / (5!*(7-5)!) = 7*6 / (2*1) = 21).

Therefore, there are 21 different combinations for placing eight identical balls into three different boxes without leaving any box empty.