Permutations and Combinations

Permutation and Combination deal with counting arrangements and selections of objects.

• Permutation = Arrangement
  → Order matters. (e.g., seating people, forming numbers, rankings)

• Combination = Selection
  → Order does not matter. (e.g., selecting a team, choosing questions)

Example:
From A, B, C:

• Permutations of 2 letters: AB, BA, AC, CA, BC, CB → 6 ways
• Combinations of 2 letters: AB, AC, BC → 3 ways

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2. Key Formulas & Shortcuts

Let $n$ = total number of items, $r$ = number to arrange or choose.

A. Factorial Notation

$$n! = n \times (n-1) \times (n-2) \times \ldots \times 1$$

By definition: $0! = 1$

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B. Permutations

1. Number of permutations of $r$ out of $n$:

$$^nP_r = \frac{n!}{(n - r)!}$$

2. All permutations of $n$ distinct items:

$$^nP_n = n!$$

3. Permutations with repetition:

If an object is repeated:

$$\text{Total permutations} = \frac{n!}{p_1! \cdot p_2! \cdot \ldots \cdot p_k!}$$

where $p_1, p_2, ..., p_k$ are counts of identical items.

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C. Combinations

1. Number of combinations of $r$ out of $n$:

$$^nC_r = \frac{n!}{r!(n - r)!}$$

2. Basic identities:

• $^nC_r = ^nC_{n-r}$
• $^nC_0 = ^nC_n = 1$

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D. Circular Permutations

• $n$ objects in a circle (no reference point):

$$(n - 1)!$$

• If clockwise and anticlockwise arrangements are considered same:

$$\frac{(n - 1)!}{2}$$

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E. Permutations with Restrictions

• If certain items must always be together:
  Treat them as one unit, then multiply by arrangements within the unit.

• If certain items must never be together:
  Total permutations – permutations with those items together.

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3. Conceptual Tips & Common Mistakes

• Permutation ≠ Combination → Watch for “arrangement” (order matters) vs “selection” (order doesn’t).
• Don’t forget to divide by factorials for repeated items.
• For “at least”/“at most” type questions, use summation over valid ranges.
• Use cases when restrictions are present.
• For circular arrangements, fix one item to break rotational symmetry.
• Understand when to use repetition allowed (e.g., forming passwords with repeated digits).

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4. Visual Explanation

Permutation vs Combination

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Circular Permutation

For 4 people A, B, C, D seated around a circular table:

Total ways = $(4-1)! = 6$
(ABC, ACB, BAC, BCA, CAB, CBA)

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5. Solved Examples

Example 1: Basic Permutation

Q: In how many ways can 3 people be seated in 5 chairs?

A:

$$^5P_3 = \frac{5!}{(5 - 3)!} = \frac{120}{2} = 60$$

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Example 2: Basic Combination

Q: From 8 players, in how many ways can you select 3 for a team?

A:

$$^8C_3 = \frac{8!}{3!5!} = 56$$

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Example 3: Permutation with Repetition

Q: How many 3-digit numbers can be formed using digits 1, 2, 3, 4 if repetition is allowed?

A:
Each place has 4 choices:

$$4 \times 4 \times 4 = 64$$

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Example 4: Permutation with Identical Items

Q: How many ways can you arrange the word “BALLOON”?

Letters: B, A, L, L, O, O, N → 7 letters
Repetition: L(2), O(2)

$$\frac{7!}{2! \cdot 2!} = \frac{5040}{4} = 1260$$

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Example 5: Circular Permutation

Q: In how many ways can 6 people sit around a circular table?

A:

$$(6 - 1)! = 5! = 120$$

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Example 6: Restriction – Always Together

Q: How many ways can A and B be always together in a row of 5 people?

A:

• Treat A+B as 1 block → now 4 people
• Ways to arrange blocks = $4!$
• Ways to arrange A and B in the block = $2!$

$$\Rightarrow 4! \times 2! = 24 \times 2 = 48$$

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Example 7: Restriction – Never Together

Q: How many ways can A and B never be together in a row of 5?

A:

• Total arrangements = $5! = 120$
• A and B together = $4! \times 2! = 48$

$$\Rightarrow \text{Not together} = 120 - 48 = 72$$