Welcome to Probability & Combinatorics
Welcome to your study notes for Probability in CCEA AS Level Further Mathematics (Unit AS 2: Section C Statistics). Don't worry if counting techniques and probability calculations feel daunting at first—once you master a few intuitive strategies, you will find these problems deeply satisfying and systematic!
In this chapter, we explore how to count outcomes efficiently without listing them all by hand. We will move from basic counting principles to permutations (arrangements where order matters) and combinations (selections where order does not matter), before using these tools to solve complex probability problems.
1. Fundamental Counting Principles
Before jumping into formulas, let's look at the two golden rules of counting that underpin everything in this unit.
The Multiplication Rule of Counting
If one task can be completed in \(m\) distinct ways, and a second subsequent task can be completed in \(n\) distinct ways, then both tasks in sequence can be performed in \(m \times n\) distinct ways.
Analogy: Imagine choosing an outfit. If you have \(3\) shirts and \(4\) pairs of trousers, you have \(3 \times 4 = 12\) different outfit combinations.
The Addition Rule of Counting
If two tasks are mutually exclusive (meaning they cannot happen at the same time), and task A can be done in \(m\) ways while task B can be done in \(n\) ways, then choosing to do task A or task B can be done in \(m + n\) ways.
Analogy: If you decide to travel home by bus (where there are \(3\) different bus routes) or by train (where there are \(2\) different train lines), you have \(3 + 2 = 5\) distinct travel options.
Key Takeaway: Remember the general rule of thumb: AND means Multiply (\(\times\)), while OR means Add (\(+\)).
2. Factorials, Permutations, and Combinations
Factorial Notation
For any positive integer \(n\), the factorial symbol \(n!\) represents the product of all positive integers from \(n\) down to \(1\):
\(n! = n \times (n - 1) \times (n - 2) \times \dots \times 2 \times 1\)
By mathematical definition, we also have:
\(0! = 1\)
Example: \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\).
Permutations (Arrangements Where Order Matters)
A permutation is an arrangement of items in a specific order. When the order of placement changes the outcome (such as assigning distinct positions like 1st, 2nd, and 3rd place, or arranging letters in a row), we are dealing with permutations.
1. Arranging \(n\) distinct objects in a line:
The total number of ways to arrange \(n\) distinct objects is:
\(n!\)
2. Arranging \(r\) objects chosen from \(n\) distinct objects:
When we choose and order \(r\) objects from a pool of \(n\) distinct items, we use the formula:
\({}^n\text{P}_r = \frac{n!}{(n - r)!}\)
Example: How many ways can \(3\) runners finish in 1st, 2nd, and 3rd place out of \(8\) competitors?
\({}^8\text{P}_3 = \frac{8!}{(8 - 3)!} = \frac{8!}{5!} = 8 \times 7 \times 6 = 336\) ways.
3. Permutations with Repeated Elements:
If we arrange \(n\) items where \(p\) items are identical of one type, \(q\) are identical of another type, and \(r\) are identical of another type, we divide by the factorials of the repeated counts to avoid double-counting identical arrangements:
\(\frac{n!}{p!\,q!\,r!}\)
Example: Find the number of arrangements of all the letters in the word STATISTICS.
Total letters \(n = 10\).
Identical letters: \(3\) S's, \(3\) T's, \(2\) I's, \(1\) A, \(1\) C.
Number of arrangements = \(\frac{10!}{3!\,3!\,2!\,1!\,1!} = \frac{3628800}{6 \times 6 \times 2} = 50400\).
Combinations (Selections Where Order Does Not Matter)
A combination is a selection of items where the order of choosing does not matter (for example, picking a committee of \(3\) people, or choosing \(5\) lottery numbers).
The number of ways to select \(r\) items from \(n\) distinct items is denoted by \({}^n\text{C}_r\) or \(\binom{n}{r}\):
\({}^n\text{C}_r = \binom{n}{r} = \frac{n!}{r!(n - r)!}\)
Symmetry Property: Notice that choosing \(r\) items to include is identical to choosing \(n - r\) items to leave behind:
\(\binom{n}{r} = \binom{n}{n - r}\)
Example: Selecting \(2\) committee members from a group of \(7\) candidates:
\({}^7\text{C}_2 = \frac{7!}{2!\,5!} = \frac{7 \times 6}{2 \times 1} = 21\) ways.
Quick Review Box: How to Choose Between \({}^n\text{P}_r\) and \({}^n\text{C}_r\)
Ask yourself: Does the order matter?
• If YES (e.g., words, seated rows, ranked roles) \(\rightarrow\) Use Permutations (\({}^n\text{P}_r\)).
• If NO (e.g., groups, teams, committees, hand of cards) \(\rightarrow\) Use Combinations (\({}^n\text{C}_r\)).
3. Standard Arrangement Restrictions & Problem Types
Examiners frequently add restrictions to arrangement and selection problems. Below are the standard methods you need to know for CCEA AS Level Further Mathematics.
Method 1: The "Block" or "Glue" Method (Items Kept Together)
When specific items must stay together, treat them as a single combined "block".
Step 1: Glue the constrained items into one single item.
Step 2: Count the total number of items to arrange (the remaining loose items + the single block), and calculate the arrangements.
Step 3: Multiply by the number of internal arrangements within the glued block.
Worked Example: In how many ways can \(4\) boys and \(3\) girls be arranged in a line if all \(3\) girls must sit together?
• Treat the \(3\) girls as \(1\) block.
• We now arrange \(4\) boys + \(1\) block = \(5\) entities. This can be done in \(5!\) ways.
• The \(3\) girls can be arranged internally among themselves in \(3!\) ways.
• Total arrangements = \(5! \times 3! = 120 \times 6 = 720\).
Method 2: The "Gap" or "Insertion" Method (Items Kept Separate)
When items must not be next to each other (i.e. strictly separated), do not try to arrange everyone at once!
Step 1: Arrange all the unrestricted items in a line first.
Step 2: Identify the "gaps" created around and between these unrestricted items (including the two outer ends). If there are \(m\) unrestricted items, there are always \(m + 1\) available gaps.
Step 3: Place the restricted items into these separate gaps using permutations.
Worked Example: In how many ways can \(5\) mathematics books and \(3\) physics books be placed on a shelf if no two physics books are together?
• First, arrange the \(5\) mathematics books: \(5!\) ways.
• This creates \(5 + 1 = 6\) available gaps: _ M _ M _ M _ M _ M _
• We need to place the \(3\) physics books into these \(6\) distinct gaps: \({}^6\text{P}_3\) ways.
• Total arrangements = \(5! \times {}^6\text{P}_3 = 120 \times (6 \times 5 \times 4) = 120 \times 120 = 14400\).
Method 3: Complementary Counting (The Subtraction Method)
Often, directly calculating a condition (such as "at least one") requires adding up many separate cases. Instead, subtract the unwanted cases from the total unrestricted count:
Number of valid ways = Total unrestricted ways \(-\) Number of violating ways
Example Formula for "At least 1":
\(N(\text{at least 1}) = N(\text{Total}) - N(\text{None})\)
Method 4: Conditional Group Selections (Forming Committees)
When selecting a committee with specific requirements from distinct sub-groups, calculate the combinations for each sub-group separately and multiply them together using the Multiplication Principle.
Worked Example: A committee of \(5\) people is to be chosen from \(6\) men and \(8\) women. How many committees can be formed if the committee must contain exactly \(2\) men and \(3\) women?
• Ways to choose \(2\) men from \(6\): \(\binom{6}{2} = 15\)
• Ways to choose \(3\) women from \(8\): \(\binom{8}{3} = 56\)
• Total possible committees = \(\binom{6}{2} \times \binom{8}{3} = 15 \times 56 = 840\).
4. Applying Combinatorics to Classical Probability
Equally Likely Outcomes
When all outcomes in a finite sample space \(S\) are equally likely, the probability of an event \(E\) occurring is given by:
\(P(E) = \frac{n(E)}{n(S)}\)
where \(n(E)\) is the number of favorable arrangements or selections, and \(n(S)\) is the total number of possible arrangements or selections.
Crucial Rule: Both the numerator \(n(E)\) and the denominator \(n(S)\) must be calculated using the same counting framework (either both using combinations, or both using ordered permutations).
Selection Probability Without Replacement (Hypergeometric Model)
Suppose a total population of size \(N\) contains \(K\) items of a particular type (successes) and \(N - K\) items of another type (failures). If a random sample of size \(n\) is chosen without replacement, the probability of selecting exactly \(k\) successes is:
\(P(X = k) = \frac{\binom{K}{k}\binom{N - K}{n - k}}{\binom{N}{n}}\)
Comprehensive Probability Example:
A box contains \(7\) red counters and \(5\) blue counters (total \(12\) counters). Four counters are selected at random without replacement. Find the probability that:
(a) Exactly \(3\) red counters and \(1\) blue counter are selected:
• Favorable selections \(n(E) = \binom{7}{3} \times \binom{5}{1} = 35 \times 5 = 175\)
• Total possible selections \(n(S) = \binom{12}{4} = 495\)
• \(P(\text{3 red, 1 blue}) = \frac{175}{495} = \frac{35}{99}\)
(b) At least \(1\) blue counter is selected:
Using complementary probability:
• \(P(\text{at least 1 blue}) = 1 - P(\text{no blue counters})\)
• "No blue counters" means all \(4\) selected are red: \(n(\text{all red}) = \binom{7}{4} = 35\)
• \(P(\text{no blue}) = \frac{\binom{7}{4}}{\binom{12}{4}} = \frac{35}{495} = \frac{7}{99}\)
• \(P(\text{at least 1 blue}) = 1 - \frac{7}{99} = \frac{92}{99}\)
5. Common Pitfalls & Examiner Warnings
Be on your guard against these classic mistakes frequently highlighted in CCEA examiner reports:
• Mixing Permutations and Combinations in Probabilities: Never use \({}^n\text{P}_r\) for the numerator and \({}^n\text{C}_r\) for the denominator. If your sample space counts unordered sets, your event must also count unordered sets.
• Forgetting Internal Rearrangements in the Block Method: Students often correctly treat a group of items as \(1\) unit, but forget to multiply by the factorial of the items inside that unit (e.g. forgetting the \(k!\) internal permutations).
• Miscounting Available Gaps: Remember that placing \(m\) items in a row creates \(m + 1\) available gaps (including the ends), not \(m\) gaps.
• Double Counting Overlapping Cases: When breaking a problem into separate cases, make sure the cases are strictly mutually exclusive so that no arrangement is counted twice.
Chapter Summary Checklist
Before moving on to practice exam questions, ensure you can confidently:
• State and apply the Multiplication and Addition principles.
• Calculate factorials, permutations \({}^n\text{P}_r\), and combinations \({}^n\text{C}_r\).
• Handle repeated elements using \(\frac{n!}{p!\,q!\,r!}\).
• Solve "together" problems using the Block Method.
• Solve "separated" problems using the Gap Method with \(m + 1\) slots.
• Apply complementary counting for "at least one" problems.
• Calculate probabilities of selections without replacement using \(P(E) = \frac{n(E)}{n(S)}\).