Slot Filling Rule

Counting Choices by Slots

The slot filling rule counts arrangements by multiplying the number of valid choices for each successive slot. It applies when every complete outcome can be built through a fixed sequence of choices.

A form with several fields follows the same pattern. Each field is a slot and each allowed entry is a choice. Multiplying the valid choices gives the number of complete forms.

Listing Choices in a Table

The table method lists every combination. Rows represent choices from one category and columns represent choices from another.

Suppose a student wants to choose an online learning package. There are three platforms (Platform A, Platform B, Platform C) and four subjects (Mathematics, Physics, Chemistry, Biology).

A table lists all possible combinations:

Tree Diagram Method

In a tree diagram, each branch records one choice. A complete path from the starting point to an endpoint represents one complete outcome.

For the same package selection, the first branches choose a platform and the next branches choose a subject.

1. Level 1:

   From $1$ starting point, branch into $3$ platform choices.

2. Level 2:

   $$\begin{aligned} &3 \text{ platforms} \times 4 \text{ subjects} \\ &= 12 \text{ endpoints} \end{aligned}$$

3. Branch structure:

   Start by choosing a platform, then choose a subject.

4. Total complete paths:

   $$3 \times 4 = 12 \text{ combinations}$$

Multiplication Rule Method

The multiplication rule gives the same count without listing every outcome. If there are $n$ slots and slot $i$ has $k_i$ valid choices, then the number of complete outcomes is:

Counting Access Codes

A school wants to create access codes for digital learning systems. The code consists of:

• First slot: $3 \text{ letters}$ ($A, B, C$)
• Second slot: $5 \text{ numbers}$ ($1, 2, 3, 4, 5$)
• Third slot: $2 \text{ symbols}$ ($@, \#$)

Multiply the choices for the three slots:

Cases with Restrictions

Restrictions can change the number of choices from one slot to the next. Decide for each slot whether an earlier choice removes an option, because that is what decides whether the factors stay equal. The examples below work through both cases.

Repetition Not Allowed

If the same object cannot be used repeatedly, then each filled slot will reduce the choices for the next slot.

Example: Creating a $3 \text{-digit}$ number from digits $2, 3, 4, 5, 6$ without repetition.

The first slot has $5$ choices. Once $1$ digit has been used, $4$ choices remain for the second slot. After $2$ digits have been used, the third slot has $3$ choices.

Repetition Allowed

If the same object can be used repeatedly, then the choices in each slot remain the same.

For the same case with repetition allowed:

Exercises

Each problem fills several places in order. Count the choices for one place at a time, multiply them, and watch for the problem that forbids repetition.

1. An electronics store sells smartphones with $4$ different brands, each available in $3$ memory capacities and $5$ color choices. How many different smartphone combinations are there?

2. A password contains $1$ letter followed by $2$ digits. Choose the letter from A, B, C, D and the digits from $\{1, 2, 3, 4, 5\}$ without repetition. How many passwords are possible?

3. From city $P$ to city $R$ through city $Q$, there are $3$ roads from $P$ to $Q$ and $4$ roads from $Q$ to $R$. How many different routes can be chosen for the journey from $P$ to $R$?

4. A license plate contains $2$ letters followed by $3$ digits. If all $26$ letters and the $10$ digits $0\text{-}9$ are available and repetition is allowed, how many plates are possible?

Answer Key

1. Given: $4 \text{ brands}$, $3 \text{ memory capacities}$, $5 \text{ color choices}$

   The number of smartphone combinations is:

   $$4\times3\times5=60$$

2. Given: $1 \text{ letter}$ from $\{A, B, C, D\}$, $2 \text{ digits}$ from $\{1, 2, 3, 4, 5\}$ without repetition

   There are $4$ choices for the letter and $5$ for the first digit. Repetition is forbidden, so only $4$ choices remain for the second digit.

   $$\begin{aligned} &\text{Total passwords} \\ &= 4 \times 5 \times 4 \\ &= 80 \text{ passwords} \end{aligned}$$

3. Given: $3 \text{ roads}$ from $P$ to $Q$, $4 \text{ roads}$ from $Q$ to $R$

   $$\begin{aligned} &\text{Total routes} \\ &= 3 \times 4 \\ &= 12 \text{ different routes} \end{aligned}$$

4. Given: $2 \text{ letters}$ from $26 \text{ letters}$, $3 \text{ digits}$ from $10 \text{ digits}$, repetition allowed

   $$\begin{aligned} &\text{First letter choices} = 26 \end{aligned}$$

   The second letter has $26$ choices.

   $$\text{Choices for each digit} = 10$$

   The total number of license plates is:

   $$26\times26\times10\times10\times10$$

   $$\begin{aligned} &= 676 \times 1000 \\ &= 676000 \text{ license plates} \end{aligned}$$