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Permutations - 10 Practice Questions

Understanding Permutations

10 Practice Questions - Learn from Every Answer!

Question 1 of 10

Basic Factorial Arrangements

In how many different ways can 5 different books be arranged on a shelf?

📚 Complete Solution:

  • Type: Basic arrangement of n distinct objects
  • Formula: n! (n factorial)
  • Given: n = 5 books
  • Calculate: 5! = 5 × 4 × 3 × 2 × 1
  • Step by step: 5 × 4 = 20; 20 × 3 = 60; 60 × 2 = 120; 120 × 1 = 120

✓ Correct Answer: D) 120 ways

Question 2 of 10

Selecting and Arranging (nPr)

In a race with 7 runners, how many different ways can Gold, Silver, and Bronze medals be awarded?

📚 Complete Solution:

  • Type: Permutation - selecting r from n objects
  • Formula: nPr = n!/(n-r)!
  • Given: n = 7 runners, r = 3 medals
  • Calculate: 7P3 = 7!/(7-3)! = 7!/4!
  • Simplify: 7P3 = 7 × 6 × 5 = 210

✓ Correct Answer: C) 210 ways

Question 3 of 10

Circular Permutations

In how many ways can 5 people be seated around a circular table?

📚 Complete Solution:

  • Type: Circular arrangement
  • Formula: (n-1)! for circular arrangements
  • Why (n-1)? Fix one person's position to eliminate rotational duplicates
  • Given: n = 5 people
  • Calculate: (5-1)! = 4! = 4 × 3 × 2 × 1 = 24

✓ Correct Answer: A) 24 ways

Question 4 of 10

Arrangements with Restrictions (Together)

How many ways can the letters A, B, C, D, E, F be arranged if A and B must always be next to each other?

📚 Complete Solution:

  • Type: Arrangement with objects that must stay together
  • Method: Treat A and B as one unit
  • Step 1: Objects become (AB), C, D, E, F = 5 units
  • Step 2: Arrange 5 units: 5! = 120
  • Step 3: A and B can swap within their unit: 2! = 2
  • Final: 120 × 2 = 240

✓ Correct Answer: B) 240 ways

Question 5 of 10

Selecting Officers

A club has 12 members. In how many ways can they elect a President, Vice President, and Secretary (all different people)?

📚 Complete Solution:

  • Type: Permutation - selecting and arranging r from n
  • Formula: nPr = n!/(n-r)!
  • Given: n = 12 members, r = 3 positions
  • Calculate: 12P3 = 12!/(12-3)! = 12!/9!
  • Simplify: 12P3 = 12 × 11 × 10 = 1,320

✓ Correct Answer: B) 1,320 ways

Question 6 of 10

Password Creation

How many different 4-digit passwords can be created using the digits 0-9 if no digit can be repeated?

📚 Complete Solution:

  • Type: Permutation without repetition
  • Formula: nPr = n!/(n-r)!
  • Given: n = 10 digits (0-9), r = 4 positions
  • Calculate: 10P4 = 10!/(10-4)! = 10!/6!
  • Simplify: 10P4 = 10 × 9 × 8 × 7 = 5,040

✓ Correct Answer: B) 5,040 passwords

Question 7 of 10

Advanced Circular Arrangements

Six friends want to sit around a circular table. If two specific friends (Alex and Jamie) must sit next to each other, how many arrangements are possible?

📚 Complete Solution:

  • Type: Circular arrangement with restriction
  • Method: Treat Alex and Jamie as one unit
  • Step 1: Now we have 5 units to arrange in a circle
  • Step 2: Circular arrangements of 5 units: (5-1)! = 4! = 24
  • Step 3: Alex and Jamie can swap: 2! = 2
  • Final: 24 × 2 = 48

✓ Correct Answer: B) 48 ways

Question 8 of 10

Arrangements with "NOT Together" Restriction

In how many ways can 8 books be arranged on a shelf if 2 specific books must NOT be next to each other?

📚 Complete Solution:

  • Type: Arrangement with "NOT together" restriction
  • Method: Total arrangements - Arrangements where they're together
  • Total arrangements: 8! = 40,320
  • Together arrangements: Treat 2 books as 1 unit
  • Now 7 units: 7! = 5,040
  • 2 books can swap: 5,040 × 2 = 10,080
  • Final: 40,320 - 10,080 = 30,240

✓ Correct Answer: C) 30,240 ways

Question 9 of 10

Even Number Formation

How many 3-digit even numbers can be formed using the digits 1, 2, 3, 4, 5 without repetition?

📚 Complete Solution:

  • Type: Permutation with constraint (must be even)
  • Key: Number must end in 2 or 4 (even digits)
  • Step 1: Choose last digit (even): 2 choices (2 or 4)
  • Step 2: Choose first digit: 4 remaining choices
  • Step 3: Choose second digit: 3 remaining choices
  • Final: 2 × 4 × 3 = 24

✓ Correct Answer: B) 24 numbers

Question 10 of 10

Complex Multi-Step Problem

A school has 8 students. They need to form a line where the tallest student must be first, and the shortest student must be last. How many different arrangements are possible?

📚 Complete Solution:

  • Type: Arrangement with fixed positions
  • Method: Fix the restricted positions, arrange the rest
  • Step 1: Tallest is fixed in position 1: 1 way
  • Step 2: Shortest is fixed in position 8: 1 way
  • Step 3: Arrange remaining 6 students in middle: 6! ways
  • Calculate: 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

✓ Correct Answer: C) 720 ways

Remember: Every mistake is a step toward mastery! 📚✨

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