1st Year Math Chapter 7 Exercise 7.3 Notes: Permutation of Things Not All Different, and Circular Permutation (Punjab Board 2026-27)
Complete concept notes, formulas, and solved examples for ICS/FSc Part 1 Mathematics, Punjab Board (PECTAA), Single National Curriculum 2026-27 session.
Exercise 7.3 extends the permutation ideas from Exercise 7.2 to two special cases: arranging objects when some are identical, and arranging objects around a circle rather than in a row.
What Does This Exercise Cover?
This exercise teaches how repeated objects reduce the number of distinct arrangements, and how circular arrangements are counted differently from linear ones since rotations of the same arrangement aren’t counted separately.
Key Concepts
1. Permutation of Things Not All Different
When some objects in a group are identical, the number of distinct arrangements of all n objects, where p are alike of one kind, q are alike of another kind, and so on, is n! / (p! × q! × …).
2. Circular Permutation
When n distinct objects are arranged around a circle, rotating the whole arrangement doesn’t create a new one, so the number of distinct circular arrangements is (n − 1)!, rather than n!.
3. Circular Permutation with Reflection (Necklaces and Bracelets)
If flipping the arrangement over (clockwise vs counterclockwise) is also considered the same, as with a necklace or bracelet, the count is further divided by 2, giving (n − 1)! / 2.
Solved Examples
Example 1: Find the number of distinct arrangements of the letters of the word BALLOON (7 letters, with L repeated twice and O repeated twice).
Total letters: 7, with L appearing twice and O appearing twice.
Number of distinct arrangements = 7! / (2! × 2!) = 5040 / 4.
Answer: 1260 arrangements
Example 2: In how many ways can 5 people be seated around a circular table?
Circular permutations of n distinct objects: (n − 1)!.
(5 − 1)! = 4!.
Answer: 24 ways
Example 3: In how many ways can 6 different beads be arranged to form a necklace?
Circular arrangement: (6 − 1)! = 5! = 120.
Since a necklace can be flipped, divide by 2: 120 / 2.
Answer: 60 ways
Sample MCQs
1. The number of distinct arrangements of n objects where p are alike is:
a) n! b) n! / p! c) n! × p! d) n! − p!
Answer: b) n! / p!
2. The number of ways to arrange n distinct objects in a circle is:
a) n! b) (n − 1)! c) n! / 2 d) (n + 1)!
Answer: b) (n − 1)!
3. The number of distinct arrangements of the letters in BALLOON is:
a) 5040 b) 2520 c) 1260 d) 630
Answer: c) 1260
4. For a necklace (where clockwise and counterclockwise arrangements are the same), the count is:
a) (n − 1)! b) (n − 1)! / 2 c) n! / 2 d) n!
Answer: b) (n − 1)! / 2
5. The number of ways to seat 6 people around a round table is:
a) 720 b) 120 c) 360 d) 24
Answer: b) 120
Important Short Questions
- Explain why arrangements of identical objects require dividing by factorials.
- State the formula for the number of ways to arrange n distinct objects around a circle.
- Find the number of distinct arrangements of the letters in APPLE.
- Why is circular permutation different from linear permutation?
- In how many ways can 4 people be seated around a circular table?
Important Long Questions
- Find the number of distinct arrangements of the letters of the word MISSISSIPPI.
- In how many ways can 7 people be seated around a round table?
- In how many ways can 8 differently colored beads be arranged to form a bracelet, where flipping the bracelet gives the same arrangement?
- Explain the reasoning behind the formula (n − 1)! for circular permutations of n distinct objects.
How to Approach This Exercise Effectively
- For repeated-letter words, count each repeated letter carefully before dividing — missing a repeat is the most common mistake.
- Remember circular permutations use (n−1)!, not n! — fixing one object’s position removes the duplicate rotations.
- Only divide by 2 for necklace/bracelet-style problems where flipping is allowed — regular circular seating (like around a table) does not get this extra division.
- Write out a small example (like 3 or 4 objects) by hand to convince yourself why the circular formula differs from the linear one.
- Double-check word-arrangement problems by listing every distinct letter and its repeat count before applying the formula.
FAQs
Q: Why do we divide by (n − 1)! instead of n! for circular arrangements?
A: Because in a circle, rotating every object by one position doesn’t create a new arrangement, so fixing one object’s position and arranging the rest linearly accounts for this correctly.
Q: What’s the difference between a necklace/bracelet arrangement and seating people around a table?
A: A necklace or bracelet can be flipped over, making clockwise and counterclockwise arrangements identical, while seating people around a table typically cannot be ‘flipped’ the same way.
Q: How do repeated letters affect the number of arrangements of a word?
A: Each set of repeated letters reduces the total count, since swapping identical letters with each other doesn’t create a visibly new arrangement.
Q: Can this method handle repeated objects in a circular arrangement too?
A: Yes — the two ideas can be combined, dividing by both (n−1)! for the circular arrangement and by the repeated-object factorials, though such combined problems are less common at this level.
Notes prepared for Punjab Board (PECTAA) 1st Year Mathematics, 2026-27 session, Chapter 7: Permutations and Combinations, Exercise 7.3.