Permutations

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Permutations — Quick Facts

Definition: A permutation is an arrangement of items where order matters.

Fundamental Counting Principle: If task 1 can be done in $m$ ways and task 2 in $n$ ways, both can be done in $m \times n$ ways.

Permutation Formula: $$P(n, r) = \frac{n!}{(n-r)!} = n(n-1)(n-2)…(n-r+1)$$

Special Cases:

• $P(n, n) = n!$ (arranging all $n$ items)
• $P(n, 0) = 1$
• $P(n, 1) = n$

Circular Permutations:

• Arranging $n$ people around a table: $(n-1)!$
• If clockwise vs anticlockwise don’t matter (necklaces): $\frac{(n-1)!}{2}$

⚡ CAT Exam Tip: Use permutations when the question uses words like “arrange,” “order,” “position,” or “different ways.” Use combinations when it says “choose,” “select,” or “group.”

---

🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding.

Permutations — Study Guide

Worked Examples:

Example 1: How many 3-digit numbers can be formed from {1, 2, 3, 4, 5} if digits cannot repeat?

$P(5, 3) = \frac{5!}{(5-3)!} = \frac{5!}{2!} = 5 \times 4 \times 3 = 60$

Example 2: In how many ways can 4 people be seated in a row?

$P(4, 4) = 4! = 24$

Example 3: In how many ways can 4 people be seated around a circular table?

$(4-1)! = 3! = 6$

Permutations with Repetition:

If $n$ items include $p$ identical items of type 1, $q$ identical items of type 2, etc.: $$\frac{n!}{p! \times q! \times …}$$

Example: Arrange the letters of “BANANA”

Total letters = 6 A appears 3 times, N appears 2 times, B appears 1 time

Number of arrangements = $\frac{6!}{3! \times 2! \times 1!} = \frac{720}{12} = 60$

Arrangements with Restrictions:

Example: How many 4-digit numbers can be formed from {1, 2, 3, 4, 5} if: a) No restriction b) Must be even c) No digit repeated and must be greater than 3000

a) $5 \times 5 \times 5 \times 5 = 625$ (repetition allowed) b) Last digit even (2, 4): $5 \times 5 \times 5 \times 2 = 250$ c) First digit (3, 4, 5) = 3 choices Remaining: 4, 3, 2 choices in order = $3 \times 4 \times 3 \times 2 = 72$

⚡ Common Student Mistake: Forgetting that the leading digit cannot be zero when forming numbers. Always check for this restriction in the problem.

---

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Permutations — Comprehensive Notes

Selected Arrangements:

Example: From 6 men and 4 women, form a committee of 5 with the chairman (chairperson) being a man. In how many ways?

Step 1: Choose the chairman from 6 men: $P(6, 1) = 6$ Step 2: Choose remaining 4 from 9 people (5 remaining men + 4 women): $P(9, 4)$

Wait — the problem says “committee of 5 with chairman being a man.” This means:

• Select 4 additional members after selecting the chairman
• The remaining 4 can be any gender

Actually, the more careful approach:

• Select chairman from 6 men: 6 ways
• Select remaining 4 from 9 (since one man is already chairman, and 4 women): $C(9, 4) = 126$
• Total: $6 \times 126 = 756$

But if the chairman is a specific role that changes the permutation:

• Choose chairman: 6 options
• Choose 4 others from 9: $C(9, 4) = 126$
• Total: $6 \times 126 = 756$

Derangements with Repetition:

Example: In how many ways can the letters of “ERROR” be arranged?

Total letters: 5 (R appears 2 times, O appears 1 time, E appears 1 time, but wait… let me check: E, R, R, O, R — actually 3 Rs, 1 E, 1 O)

Wait “ERROR” has E, R, R, O, R — 5 letters with 3 Rs. Actually that’s R, R, R, E, O. So 3 Rs.

$\frac{5!}{3!} = 20$ arrangements

Seating with Constraints:

Example: 5 boys and 4 girls seated in a row. No two girls sit together. How many arrangements?

Step 1: Arrange 5 boys: $5! = 120$ Step 2: There are 6 “gaps” (_ B _ B _ B _ B _ B _) where girls can be placed Step 3: Choose 4 of these 6 gaps and arrange 4 girls: $P(6, 4) = 6 \times 5 \times 4 \times 3 = 360$

Total: $120 \times 360 = 43,200$

Arrangement of “Must Be Together”:

Example: In how many ways can the letters of “FATHER” be arranged if A and T must always be together?

Treat “AT” as one unit. Arrange: F, H, E, R, (AT) = 5 items = $5! = 120$ Within the AT unit: A then T, or T then A = $2! = 2$

Total: $120 \times 2 = 240$

Arrangement of “Never Together”:

Example: In how many ways can 5 men and 4 women be seated in a row if no two women sit together?

First arrange 5 men: $5! = 120$ There are 6 gaps: _ M _ M _ M _ M _ M _ Choose 4 gaps for 4 women: $P(6, 4) = 360$

Total: $120 \times 360 = 43,200$

JAMB Pattern Analysis (CAT 2015-2024):

• 2015: $P(n, r)$ calculation with no restrictions
• 2017: Circular permutations
• 2019: Arrangements with repeated letters
• 2021: “Must sit together” constraint
• 2023: “Never together” constraint with gaps method
• 2024: Mixed constraints — leader + committee selection

⚡ Exam Strategy: For “at least” or “never” problems, use the gaps method: arrange the fixed group first, count the gaps, then arrange the other group into selected gaps.