Question

How many ways can five identical advertisements be placed in three mailboxes if each mailbox receives at least one advertisement? How many ways if a mailbox may receive none? (The order in which a messenger delivers a message is immaterial).

Short answer

6 ways if each gets at least one; 21 ways if some can be empty.

Step-by-step solution

Understand the Problem

We have five identical advertisements that need to be distributed among three mailboxes. In the first part, each mailbox must receive at least one advertisement, while in the second part, mailboxes might receive none.

Distribute Advertisements with Non-Empty Mailboxes

Each mailbox must receive at least one advertisement. Give one advertisement to each mailbox, using 3 out of the 5. Now, distribute the remaining 2 advertisements freely among the 3 mailboxes. This becomes a problem of distributing 2 identical items into 3 different groups.

Apply Stars and Bars Method

For distributing 2 advertisements freely among 3 mailboxes (each already having 1), use the "stars and bars" combinatorics theorem. Here, number of items (advertisements) equals 2, and the number of groups (mailboxes) equals 3, so we need to find the number of non-negative integer solutions to the equation: \[ x_1 + x_2 + x_3 = 2 \]This is given by the formula \( \binom{n+k-1}{k-1} \), where \( n \) is the total number of items (2) and \( k \) is the number of groups (3).

Calculate Combinations for Any Distribution

Substitute values into the formula mentioned in Step 3: \[ \binom{2+3-1}{3-1} = \binom{4}{2} = 6 \]Thus, there are 6 ways to distribute the advertisements such that each mailbox receives at least one.

Distribute Advertisements Where None Might Be Empty

Now, allow mailboxes to potentially be empty. Hence, we have 5 identical advertisements to distribute across 3 mailboxes freely. Again use stars and bars for the non-negative integer solutions to:\[ x_1 + x_2 + x_3 = 5 \]The formula used is \( \binom{n+k-1}{k-1} \), with \( n = 5 \) and \( k = 3 \).

Calculate Combinations for Empty-Allowed Distribution

Substitute the values into the formula: \[ \binom{5+3-1}{3-1} = \binom{7}{2} = 21 \]So there are 21 ways to distribute the advertisements when some mailboxes might remain empty.

Key concepts

Stars and Bars Method

Understanding the "stars and bars" method can be a game-changer in solving combinatorics problems involving the distribution of identical objects into distinct groups. This method essentially provides a way to find the number of possible distributions of a given number of identical items into distinct groups, where some groups could potentially remain empty. Here's how it works:

The idea is to represent each identical item as a "star" and then use "bars" to separate these stars into different groups. If we have 'n' items and want to distribute them into 'k' groups, we need 'k-1' bars. The key is to determine how different arrangements of these stars and bars can occur. For example, distributing 2 stars into 3 groups would appear like this: **|**|**. This graphic notation clearly illustrates how stars are separated into groups by bars.

The classic formula used is \( \binom{n+k-1}{k-1} \), where 'n' is the number of stars, and 'k-1' represents the bars needed to divide items into groups. This formula calculates the number of ways to arrange the stars and bars linearly. Such visual methods, like stars and bars, make abstract combinatorics problems much more tangible and easier to tackle.

Integer Solutions

Integer solutions involve finding non-negative whole number solutions to equations involving multiple variables. In the context of distributing items into groups, each solution represents a way to distribute the items under given constraints.

For instance, imagine needing to find solutions to the equation \( x_1 + x_2 + x_3 = 2 \) when using the stars and bars method. Each variable \( x_i \) represents the number of items in a given group. The task here is to find all possible ordered combinations of these non-negative integers that sum up to the total number of items, 'n'.

• Non-Negative Solutions: Solutions that are equal to or greater than zero.
• Ordered Combinations: Different sequence arrangements matter, such as \((1,1,0)\) differs from \((0,1,1)\).

Finding integer solutions helps translate the abstract concept of distribution into a numerical script, making these problems more accessible. These solutions depict every feasible way to allocate resources (or items) across different categories or groups.

Distribution Problems

Distribution problems in combinatorics revolve around strategically allocating identical or distinct items into distinct groups based on specified conditions. These conditions might include allowing groups to be empty or ensuring each group gets at least one item.

Take a scenario where you have 5 identical advertisements and 3 mailboxes, where each mailbox must receive at least one advertisement. Initially, ensure each mailbox gets one, leaving you with 2 additional advertisements to distribute without restrictions. This becomes a typical stars and bars scenario.

Key aspects of distribution problems include:

• Restrictions: Constraints such as a minimum or maximum number of items per group, or whether empty groups are allowed.
• Identical vs. Distinct Items: The approach can differ if items or groups differ in nature, affecting how solutions are formed.

Using combinatorics strategies and methods like stars and bars simplifies solving these distribution problems, making complex allocations straightforward by breaking them into manageable parts. It transforms seemingly complicated distribution tasks into mathematical equations that are logically and efficiently solvable.