Question

How many ways are there to distribute five balls into three boxes if each box must have at least one ball in it if

a) both the balls and boxes are labeled?

b) the balls are labeled, but the boxes are unlabeled?

c) the balls are unlabeled, but the boxes are labeled?

d) both the balls and boxes are unlabeled?

Short answer

\({\rm{60 + 90 = 150}}\)

a) The solution is .

b) The solution is \({\rm{10 + 5 \times 3 = 25}}\).

c) The solution is \({\rm{3 + 3 = 6}}\).

d) There are only the two options.

Step-by-step solution

Concept Introduction

Counting is the act of determining the quantity or total number of objects in a set or a group in mathematics. To put it another way, to count is to say numbers in sequence while giving a value to an item in a group on a one-to-one basis. Objects are counted using counting numbers.

How many ways are there to distribute five balls into three boxes 

To begin, observe that each box must have at least one ball, therefore there are only two fundamental arrangements: three balls in one box and one ball in each of the other two boxes (denoted 3-l-1), or one ball in one box and two balls in each of the other two boxes (denoted 2-l-2) (denoted 1-2-2).

Both the balls and boxes are labeled?

1. There are three methods to choose the congested box, \({\rm{C(5,3) = 10}}\)ways to choose the balls to be placed there, and two ways to decide where the other balls go in the \({\rm{3 - 1 - 1}}\)arrangement, for a total of \({\rm{3 \times 10 \times 2 = 60}}\)options. There are \({\rm{3 \times 5 \times 6 = 90}}\)ways to choose the box that will have only one ball, 5 ways to choose which ball goes there, and \({\rm{C(4,2) = 6}}\)ways to choose which two balls go into the lower-numbered remaining box in the \({\rm{1 - 2 - 2}}\)configuration.

As a result, the solution is \({\rm{60 + 90 = 150}}\).

Step 4: The balls are labeled, but the boxes are unlabeled?

b. In the \({\rm{3 - 1 - 1}}\)configuration, there are \({\rm{C(5,3) = 10}}\)options to choose the balls for the congested box. There are five ways to choose the lonely ball in the \({\rm{1 - 22}}\)configuration, and three ways to choose the mate of the lowest-numbered remaining ball.

As a result, the solution is \({\rm{10 + 5 \times 3 = 25}}\).

The balls are unlabeled, but the boxes are labeled?

c. For the \({\rm{3 - 1 - 1}}\)arrangement, there are three ways to choose the crowded box, and for the \({\rm{1 - 2 - 2}}\)arrangement, there are three ways to choose the solo box.

As a result, the solution is \({\rm{3 + 3 = 6}}\).

Both the balls and boxes are unlabeled?

d) There are only the two options we've discussed so far: \({\rm{3 - 1 - 1}}\) and\({\rm{1 - 2 - 2}}\).