Permutation & Combination: Fundamentals, Problems & Applications

TAKEAWAYS:

• Understand the difference between permutations (order matters) and combinations (order doesn’t matter) using real-life examples like passwords vs. pizza toppings.
• Master factorial notation (n!) and its role in counting arrangements, with shortcuts for calculations.
• Apply permutation formulas (nPr) for distinct and repeated items, and combination formulas (nCr) for selections.
• Solve word problems involving circular arrangements, committee formation, and probability using these concepts.
• Recognize common NEB question patterns like "how many ways can X be arranged?" or "how many teams can be formed?"
• Avoid common mistakes like misapplying formulas or ignoring restrictions (e.g., identical items or repeated elements).

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What Are Permutations and Combinations?

Permutations and combinations are two fundamental counting techniques used to determine the number of possible arrangements or selections in a given scenario.

Permutation (Arrangement)

• Definition: The number of ways to arrange a set of items where order matters.
• Example: Arranging letters in "CAT" as "CAT," "CTA," "ACT," etc.
• Key Idea: "ABC" is different from "BAC" because the order changes the meaning.

Combination (Selection)

• Definition: The number of ways to select items where order does not matter.
• Example: Choosing 2 fruits from {apple, banana, cherry} as {apple, banana} is the same as {banana, apple}.
• Key Idea: "AB" is identical to "BA" in combinations.

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Factorial Notation (n!)

• Definition: The product of all positive integers from 1 to n. $n!=n\times (n-1)\times (n-2)\times \cdots \times 1$
• Examples:
  • $4!=4\times 3\times 2\times 1=24$
  • $0!=1$ (by definition)
• Why It Matters: Factorials are the building blocks of permutation and combination formulas.

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Permutation Formulas

There are two main types of permutations:

1. Permutation of Distinct Items (nPr)

• Formula: $P(n,r)=\frac{n!}{(n-r)!}$ (where $r\le n$)
• When to Use: When all items are unique, and order matters.
• Example: How many ways can 3 books be arranged on a shelf from 5 books? $P(5,3)=\frac{5!}{(5-3)!}=\frac{120}{2}=60$

2. Permutation of Repeated Items

• Formula: $\frac{n!}{n_1!\times n_2!\times \cdots \times n_k!}$ (where $n_1,n_2,\dots ,n_k$ are the counts of identical items)
• When to Use: When some items are identical (e.g., "MISSISSIPPI").
• Example: How many distinct arrangements are there in "MISSISSIPPI"?
  • Total letters: 11
  • Repeated letters: 4 S’s, 4 I’s, 2 P’s
  • Formula: $\frac{11!}{4!\times 4!\times 2!}=34,650$

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Combination Formulas

Combinations ignore order, so the formula is simpler:

Combination of Distinct Items (nCr)

• Formula: $C(n,r)=\frac{n!}{r!(n-r)!}$ (where $r\le n$)
• When to Use: When selecting items where order doesn’t matter (e.g., teams, committees).
• Example: How many ways can 3 students be chosen from 10? $C(10,3)=\frac{10!}{3!\times 7!}=120$

Combination with Restrictions

• Example: How many ways can a committee of 3 be formed from 5 boys and 4 girls if it must include at least 1 girl?
  • Total ways: $C(9,3)=84$
  • Unwanted ways (all boys): $C(5,3)=10$
  • Valid ways: $84-10=74$

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Special Cases and Applications

1. Circular Permutations

• When to Use: Arranging items in a circle (e.g., seating around a table).
• Formula: $(n-1)!$ (since rotations are identical)
• Example: How many ways can 4 people sit around a round table? $(4-1)!=6$ ways.

2. Permutation vs. Combination in Probability

• Example: Probability of drawing 2 kings from a deck of 52 cards.
  • Total ways to choose 2 cards: $C(52,2)=1326$
  • Favorable ways (2 kings): $C(4,2)=6$
  • Probability: $\frac{6}{1326}=\frac{1}{221}$

3. Real-Life Applications

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Solved Examples (NEB Style)

Example 1: Permutation of Distinct Items

Question: In how many ways can the letters of the word "LEADING" be arranged so that the vowels come together? Solution:

1. Treat vowels (E, A, I) as a single entity: "LE(DING)" → 5 items.
2. Arrange these 5 items: $5!=120$ ways.
3. Arrange the 3 vowels internally: $3!=6$ ways.
4. Total arrangements: $120\times 6=720$.

Example 2: Combination with Restrictions

Question: A box contains 5 red and 4 blue balls. In how many ways can 3 balls be selected such that at least one is red? Solution:

1. Total ways to choose 3 balls: $C(9,3)=84$.
2. Unwanted ways (all blue): $C(4,3)=4$.
3. Valid ways: $84-4=80$.

Example 3: Circular Permutation

Question: In how many ways can 5 people be seated around a round table if two specific people must sit together? Solution:

1. Treat the two people as a single entity → 4 entities.
2. Circular arrangements: $(4-1)!=6$ ways.
3. The two people can switch places: $6\times 2=12$ ways.

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Common Mistakes to Avoid

1. Ignoring Order: Using combinations when order matters (e.g., passwords).
2. Incorrect Factorial Calculation: Forgetting $0!=1$ or misapplying $(n-r)!$.
3. Overcounting in Circular Permutations: Not subtracting 1 for rotations.
4. Forgetting Restrictions: Not accounting for identical items or constraints (e.g., "at least one girl").
5. Mixing Formulas: Confusing $P(n,r)$ and $C(n,r)$.

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Exam Tip: How to Score Full Marks

1. Show All Steps: NEB examiners reward detailed working. Always write:
   • The formula used.
   • Substituted values.
   • Simplification steps.
2. Label Your Answer: Write "Permutation" or "Combination" clearly at the start.
3. Watch Units: Ensure $n$ and $r$ are correctly identified (e.g., $n=10$, $r=3$).
4. Practice Word Problems: NEB often tests real-world scenarios (e.g., "How many ways can a committee be formed?").
5. Check for Restrictions: Look for keywords like:
   • "At least one..."
   • "No two identical..."
   • "Arranged in a circle..."
6. Use Shortcuts: Memorize common values like $C(52,2)=1326$ (for card problems).

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NEB Board-Style Questions (Practice)

1. Permutation:

   • How many 4-letter passwords can be formed from the letters in "MATHEMATICS" if no letter is repeated?
   • Answer: $P(10,4)=5040$.

2. Combination:

   • A class has 12 boys and 8 girls. How many ways can a team of 5 be formed with at least 3 girls?
   • Answer: $C(8,3)\times C(12,2)+C(8,4)\times C(12,1)+C(8,5)=1140$.

3. Circular Permutation:

   • In how many ways can 6 people be seated around a table if two specific people refuse to sit together?
   • Answer: Total circular arrangements: $5!=120$. Unwanted arrangements (together): $2\times 4!=48$. Valid arrangements: $120-48=72$.

4. Repeated Items:

   • How many distinct arrangements are there in the word "INDEPENDENCE"?
   • Answer: $\frac{12!}{2!\times 3!\times 2!}=1,381,380$.

5. Probability with Combinations:

   • A bag contains 5 white and 4 black balls. What is the probability of drawing 2 white balls in succession without replacement?
   • Answer: $\frac{C(5,2)}{C(9,2)}=\frac{10}{36}=\frac{5}{18}$.

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Summary Table

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