Permutation & Combination: Definitions, Formulas, Applications

TAKEAWAYS:

• Permutations count ordered arrangements (e.g., passwords, race rankings), while combinations count unordered selections (e.g., committees, lottery draws).
• The formulas are $P(n,r)=\frac{n!}{(n-r)!}$ and $C(n,r)=\frac{n!}{r!(n-r)!}$, where $n!$ (n factorial) means $n\times (n-1)\times \cdots \times 1$.
• Factorials grow very fast: $5!=120$, $10!=3,628,800$, so simplify before multiplying.
• Business uses: scheduling, inventory, risk assessment, and probability (e.g., calculating chances of winning prizes).
• Always check if repetition is allowed (e.g., passwords with repeated letters) or not (e.g., unique employee IDs).
• NEB exams test both formula application and word-to-math translation (e.g., "how many ways to arrange 3 books out of 5?").

---

1. Factorials: The Building Blocks

Factorials ($n!$) are the product of all positive integers up to $n$. They are the foundation of permutations and combinations.

Definition

For any positive integer $n$: $$n!=n\times (n-1)\times (n-2)\times \cdots \times 1$$ By definition, $0!=1$.

Examples

Calculate the following factorials:

1. $4!=4\times 3\times 2\times 1=24$
2. $6!=6\times 5\times 4\times 3\times 2\times 1=720$
3. $0!=1$ (special case)

Why Factorials Matter

Factorials help count all possible arrangements of items. For example, if you have 3 books (A, B, C), the number of ways to arrange them is $3!=6$:

ABC, ACB, BAC, BCA, CAB, CBA

Visual: Factorial Growth

Key Observation: Factorials grow extremely fast. Even $10!$ is over 3 million! This is why simplifying fractions with factorials is crucial in permutations and combinations.

---

2. Permutations: Ordered Arrangements

Permutations count the number of ways to arrange $r$ items from a set of $n$ distinct items where order matters.

Formula

The number of permutations of $n$ items taken $r$ at a time is: $$P(n,r)=\frac{n!}{(n-r)!}$$ Why? We divide by $(n-r)!$ to cancel out the arrangements of the unused items.

Example 1: Basic Permutation

Problem: How many ways can you arrange 3 letters from the word "MATH"? Solution: Here, $n=4$ (M, A, T, H) and $r=3$. $$P(4,3)=\frac{4!}{(4-3)!}=\frac{4!}{1!}=\frac{24}{1}=24$$ Visual: All Possible Arrangements For "MATH" taken 3 at a time, one possible arrangement is:

MAT, MTH, AMT, AMH, TMA, TMH, HMA, HMT, ...

(There are 24 total, but we list a few to show the pattern.)

Example 2: Permutation with Repetition

Problem: How many 3-digit passwords can be made using the digits 1, 2, 3, 4 if repetition is allowed? Solution: Here, repetition is allowed, so each digit can be used more than once. The number of choices for each position is:

• First digit: 4 choices (1, 2, 3, 4)
• Second digit: 4 choices (repetition allowed)
• Third digit: 4 choices Total permutations = $4\times 4\times 4=4^3=64$.

Comparison Table: Permutations with/without Repetition

---

3. Combinations: Unordered Selections

Combinations count the number of ways to select $r$ items from a set of $n$ distinct items where order does not matter.

Formula

The number of combinations of $n$ items taken $r$ at a time is: $$C(n,r)=\frac{n!}{r!(n-r)!}$$ Why? We divide by $r!$ to account for the fact that the order of selection does not matter (e.g., selecting A and B is the same as B and A).

Example 1: Basic Combination

Problem: How many ways can you choose 2 fruits from a basket of 4 (apple, banana, cherry, date)? Solution: Here, $n=4$ and $r=2$. $$C(4,2)=\frac{4!}{2!(4-2)!}=\frac{24}{2\times 2}=\frac{24}{4}=6$$ Visual: All Possible Combinations The 6 combinations are:

{apple, banana}, {apple, cherry}, {apple, date},
{banana, cherry}, {banana, date}, {cherry, date}

Notice that {apple, banana} is the same as {banana, apple}, so we count it only once.

Example 2: Business Application

Problem: A company wants to form a 3-member committee from 7 employees. How many ways can this be done? Solution: Order does not matter (the committee {A, B, C} is the same as {B, A, C}), so we use combinations: $$C(7,3)=\frac{7!}{3!4!}=\frac{5040}{6\times 24}=\frac{5040}{144}=35$$ So, there are 35 possible committees.

Comparison Table: Permutations vs. Combinations

---

4. Key Properties and Shortcuts

Property 1: Symmetry in Combinations

$$C(n,r)=C(n,n-r)$$ Example: $$C(5,2)=C(5,3)=10$$ This is because choosing 2 items to include is the same as choosing 3 items to exclude.

Property 2: Sum of Combinations

$$C(n,0)+C(n,1)+C(n,2)+\cdots +C(n,n)=2^n$$ Example: For $n=3$: $$C(3,0)+C(3,1)+C(3,2)+C(3,3)=1+3+3+1=8=2^3$$

Visual: Pascal’s Triangle (Combinations)

Pascal’s Triangle shows how combinations grow. Each number is the sum of the two directly above it, and the $n$th row corresponds to the coefficients of $(a+b{)}^n$ in the Binomial Theorem (which you’ll study next!).

---

5. Business Applications

Permutations and combinations are used in real-world business scenarios:

Application 1: Scheduling

Problem: A manager has 5 tasks to assign to 3 employees. How many ways can this be done if each employee gets at least one task? Solution: This is a permutation problem where order matters (which employee gets which task). However, since tasks are distinct and employees are distinct, we use: $$P(5,3)=\frac{5!}{(5-3)!}=60$$ But if tasks are identical (e.g., 5 identical items to 3 distinct boxes), we’d use combinations with repetition.

Application 2: Inventory Selection

Problem: A store has 8 types of snacks. How many ways can a customer choose 4 snacks if the order doesn’t matter? Solution: This is a combination problem: $$C(8,4)=\frac{8!}{4!4!}=70$$

Application 3: Probability in Business

Problem: A lottery has 50 balls, and you pick 6. What is the probability of winning the jackpot (matching all 6 numbers)? Solution: Total ways to pick 6 numbers: $C(50,6)$. Only 1 winning combination. Probability = $\frac{1}{C(50,6)}=\frac{1}{15,890,700}$.

---

6. Common Mistakes to Avoid

1. Confusing Permutations and Combinations:

   • Wrong: Using $P(n,r)$ when order doesn’t matter.
   • Right: Use $C(n,r)$ for unordered selections.

2. Forgetting to Simplify Factorials:

   • Wrong: Calculating $C(10,3)=\frac{10!}{3!7!}$ as $\frac{3,628,800}{6\times 5040}$ without simplifying.
   • Right: Simplify before multiplying: $$C(10,3)=\frac{10\times 9\times 8}{3\times 2\times 1}=120$$

3. Ignoring Repetition:

   • Wrong: Assuming no repetition when it’s allowed (e.g., passwords).
   • Right: Use $n^r$ for permutations with repetition.

4. Misapplying Formulas:

   • Wrong: Using $P(n,r)$ when items are not distinct (e.g., identical balls).
   • Right: Use combinations or adjust for identical items.

---

7. Solved NEB-Style Problems

Problem 1: Permutation

Question: In how many ways can 4 people be seated in a row of 5 chairs if no two people can sit together? Solution: This is a permutation problem with restrictions. First, arrange the 4 people in 5 chairs such that no two are adjacent.

• Total ways to choose 4 chairs out of 5: $C(5,4)=5$.
• Arrange 4 people in these chairs: $4!=24$. Total arrangements = $5\times 24=120$.

Problem 2: Combination

Question: A class has 12 boys and 8 girls. How many ways can a committee of 5 be formed with at least 3 girls? Solution: We break this into cases:

1. 3 girls and 2 boys: $C(8,3)\times C(12,2)=56\times 66=3696$
2. 4 girls and 1 boy: $C(8,4)\times C(12,1)=70\times 12=840$
3. 5 girls and 0 boys: $C(8,5)\times C(12,0)=56\times 1=56$ Total ways = $3696+840+56=4592$.

Problem 3: Mixed Problem

Question: How many 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5 if: a) Repetition is not allowed? b) Repetition is allowed? Solution: a) No repetition:

• First digit: 5 choices (1-5)
• Second digit: 4 remaining choices
• Third digit: 3 choices
• Fourth digit: 2 choices Total = $5\times 4\times 3\times 2=120$ or $P(5,4)=120$.

b) With repetition:

• Each of the 4 digits has 5 choices. Total = $5^4=625$.

---

8. Exam Tip: How to Score Full Marks

1. Understand the Scenario:

   • Ask: "Does order matter?" (Permutation) or "Does it not matter?" (Combination).
   • Example: Arranging employees in a line (order matters) vs. selecting a team (order doesn’t matter).

2. Write the Formula Clearly:

   • Always write $P(n,r)$ or $C(n,r)$ before substituting values. This shows your understanding.

3. Simplify Before Multiplying:

   • Cancel common terms in factorials to avoid large calculations. For example: $$C(10,3)=\frac{10\times 9\times 8}{3\times 2\times 1}=120$$ instead of calculating $10!$ directly.

4. Show All Steps:

   • NEB examiners reward step-by-step working. Even if the answer is simple, show how you arrived there.

5. Practice Word-to-Math Translation:

   • Convert phrases like:
     • "How many ways to arrange..." → Permutation
     • "How many ways to choose..." → Combination
     • "With repetition" → Adjust formula accordingly.

6. Check Units and Reasonableness:

   • If the answer is a large number (e.g., 100,000), ask: "Does this make sense?" For example, $C(50,6)$ should be a very large number (it is: 15,890,700).

7. Memorize Key Values:

   • Know $0!=1$, $1!=1$, $2!=2$, $3!=6$, $4!=24$, $5!=120$. This speeds up calculations.

---

9. NEB Board-Style Questions for Practice

Section A: Short Answer (5 marks each)

1. Calculate $P(7,3)$ and $C(7,3)$. What do you observe?
2. In how many ways can 5 distinct books be arranged on a shelf if 2 specific books must always be together?
3. A committee of 4 is to be formed from 6 men and 4 women. How many ways can this be done if: a) There are no restrictions? b) The committee must include at least 2 women?
4. Explain why $C(n,r)=C(n,n-r)$ with an example.
5. How many 3-digit even numbers can be formed using the digits 1, 2, 3, 4, 6 if repetition is not allowed?

Section B: Long Answer (10 marks each)

1. A company has 8 employees. It wants to form a team of 4 to work on a project. However, 2 employees cannot work together due to a conflict. How many valid teams can be formed?
2. A password consists of 4 digits where:
   • The first digit cannot be 0.
   • Repetition of digits is allowed. How many such passwords are possible?
3. In a class of 30 students, 12 are girls and 18 are boys. How many ways can a president and vice-president be chosen if: a) Both must be girls? b) One must be a girl and the other a boy?
4. Prove that $C(n,0)+C(n,1)+C(n,2)+\cdots +C(n,n)=2^n$ using combinations.
5. A box contains 5 red balls and 4 blue balls. How many ways can 6 balls be selected such that there are at least 2 red balls?

---

Final Note: Permutations and combinations are fundamental to Business Mathematics. Master these concepts, and you’ll excel in probability, statistics, and even linear programming later in the syllabus. Practice daily, and don’t hesitate to draw diagrams or list small cases to understand patterns!