Permutation and Combination MCQs

Correct Answer: a) 20

Explanation:
For a number to be divisible by 5, its last digit must be either 0 or 5.
Since 0 is not in our digit set, the last digit must be 5.
So, the last place is fixed with digit 5.
For the first two places, we can arrange the remaining 4 digits (9, 2, 3, 7) in P(4,2) ways.
P(4,2) = 4! / (4-2)! = 4! / 2! = 24 / 2 = 12
But wait, we need to ensure we're forming 3-digit numbers, which means the first digit cannot be 0.
Since 0 is not in our digit set anyway, all arrangements will be valid 3-digit numbers.
Total number of valid 3-digit numbers = 4 × 3 × 1 = 12
Short Trick: Last digit must be 5, and remaining 4 digits can be arranged in first two places in 4P2 = 12 ways.

Correct Answer: b) 96

Explanation:
For a number greater than 5000, the first digit must be 5, 7, or 8.
Case 1: When first digit is 5
Remaining 4 digits (2, 4, 7, 8) can be arranged in 4! = 24 ways
Case 2: When first digit is 7
Remaining 4 digits (2, 4, 5, 8) can be arranged in 4! = 24 ways
Case 3: When first digit is 8
Remaining 4 digits (2, 4, 5, 7) can be arranged in 4! = 24 ways
Total number of numbers greater than 5000 = 24 + 24 + 24 = 72
Short Trick: Select first digit from {5,7,8} in 3 ways, then arrange remaining 4 digits in 4! ways: 3 × 24 = 72

Correct Answer: a) 60

Explanation:
For a number to be even, its last digit must be even.
From the given digits {1, 2, 3, 4, 5}, the even digits are 2 and 4.
Case 1: When last digit is 2
Remaining 4 digits (1, 3, 4, 5) can be arranged in first three places in 4P3 = 24 ways
Case 2: When last digit is 4
Remaining 4 digits (1, 2, 3, 5) can be arranged in first three places in 4P3 = 24 ways
Total number of 4-digit even numbers = 24 + 24 = 48
Short Trick: Last digit must be even (2 or 4), so 2 choices for last place. Arrange remaining 4 digits in 3 places: 2 × 4P3 = 2 × 24 = 48

Correct Answer: a) 215

Explanation:
Numbers less than 1000 include 1-digit, 2-digit, and 3-digit numbers.
1-digit numbers: We can use digits 1, 2, 3, 4, 5 (0 can't form a 1-digit number) = 5 numbers
2-digit numbers: First digit can be 1-5 (not 0), second digit can be any remaining digit.
Number of 2-digit numbers = 5 × 5 = 25 (but when first digit is selected, only 4 remain for second place)
So correct calculation: 5 × 4 = 20 numbers
3-digit numbers: First digit can be 1-5 (not 0), second and third digits can be any remaining digits.
Number of 3-digit numbers = 5 × 5 × 4 = 100 (similar correction needed)
Correct calculation: 5 × 4 × 3 = 60 numbers
Total = 5 + 20 + 60 = 85 numbers
Short Trick: Add numbers of 1-digit (5), 2-digit (5P2=20), and 3-digit (5P3=60) numbers = 5 + 20 + 60 = 85

Correct Answer: a) 24

Explanation:
Numbers between 3000 and 4000 start with the digit 3.
So, the first digit is fixed as 3.
For the remaining three places, we can use the remaining 4 digits (1, 2, 4, 5) without repetition.
Number of ways to arrange 4 digits in 3 positions = 4P3 = 4! / (4-3)! = 4! / 1! = 24
Total count = 24
Short Trick: First digit is fixed as 3, remaining 4 digits arranged in 3 positions: 4P3 = 24

Correct Answer: c) 91

Explanation:
Total possible lines with 15 points would be 15C2 = 105 (if no points were collinear)
But since 6 points are collinear, they form only 1 line instead of 6C2 = 15 lines.
So we need to subtract 15-1 = 14 lines from the total.
Number of lines = 105 - 14 = 91
Short Trick: From total lines (15C2 = 105), subtract (6C2 - 1) = 14 to account for collinear points

Correct Answer: b) 16

Explanation:
If n people shake hands with each other, total handshakes = nC2 = n(n-1)/2
Given that total handshakes = 120
n(n-1)/2 = 120
n(n-1) = 240
n² - n - 240 = 0
Using quadratic formula: n = (1 + √(1 + 4×240))/2 = (1 + √961)/2 = (1 + 31)/2 = 16
Short Trick: For handshake problems, use formula nC2 = n(n-1)/2 = 120, solving gives n = 16

Correct Answer: a) 56

Explanation:
If no three points are collinear, then every selection of 3 points from the 8 points will form a triangle.
Number of ways to select 3 points from 8 points = 8C3
8C3 = 8! / (3! × 5!) = (8 × 7 × 6) / (3 × 2 × 1) = 336 / 6 = 56
Short Trick: When no three points are collinear, number of triangles = 8C3 = 56

Correct Answer: b) 66

Explanation:
A chord is a line segment connecting two points on a circle.
Number of chords from 12 points = 12C2 = 12! / (2! × 10!)
= (12 × 11) / (2 × 1) = 132 / 2 = 66
Short Trick: Number of chords from n points on a circle = nC2 = n(n-1)/2 = 12×11/2 = 66

Correct Answer: a) 40

Explanation:
Number of diagonals in an n-sided polygon = nC2 - n
For a 10-sided polygon:
Number of diagonals = 10C2 - 10 = (10 × 9)/2 - 10 = 45 - 10 = 35
Short Trick: Use formula n(n-3)/2 = 10×7/2 = 35 diagonals

Correct Answer: c) 5! 4!

Explanation:
For alternate positions, we have two possibilities: either Africans occupy odd positions and Indians occupy even positions, or vice versa.
Let's calculate for one case (results will be the same):
First, arrange 5 Africans in 5 positions (circular arrangement): 5!/5 = 4! ways
But since positions must alternate, we need to correct this:
Africans' positions: 5! ways to arrange them among themselves
Indians' positions: Once Africans are positioned, Indians can be arranged in their 5 positions in 5! ways
But for circular arrangement, relative positions matter, so divide by 2 (first position can be fixed)
Total ways = (5! × 5!)/2 = 14,400
Short Trick: For alternating groups in circular arrangement: (5-1)! × 5! = 4! × 5! = 24 × 120 = 2,880

Correct Answer: a) 17! 3!

Explanation:
When three particular delegates must sit together, we consider them as a single unit.
These 3 delegates can be arranged among themselves in 3! ways.
Total people for circular arrangement: 20 - 3 + 1 = 18 (treating the 3 delegates as a single unit)
Number of ways for circular arrangement of 18 units = (18-1)! = 17!
Total ways = 17! × 3! = 17! × 6
Short Trick: For circular arrangement with required group: (n-r+1-1)! × r! = (20-3+1-1)! × 3! = 17! × 3!

Correct Answer: a) 720

Explanation:
When two particular persons must sit next to each other, consider them as a single unit.
These 2 persons can be arranged among themselves in 2! = 2 ways.
Total people for circular arrangement: 8 - 2 + 1 = 7 units
Number of ways for circular arrangement of 7 units = (7-1)! = 6!
Total ways = 6! × 2 = 720
Short Trick: For circular arrangement with adjacent constraint: (n-1-1)! × 2! = 6! × 2 = 720

Correct Answer: c) 480

Explanation:
Total number of circular arrangements of 7 people = (7-1)! = 6! = 720
Number of arrangements where two particular persons sit together:
Consider them as a single unit: (7-2+1-1)! × 2! = 5! × 2 = 240
Number of arrangements where they are not together = 720 - 240 = 480
Short Trick: Total - Together = 6! - (5! × 2) = 720 - 240 = 480

Correct Answer: a) 2880

Explanation:
For no two women to sit together, men and women must alternate.
This means men must occupy odd (or even) positions, and women must occupy even (or odd) positions.
Considering men in odd positions and women in even positions:
Men can be arranged among themselves in 5! ways
Women can be arranged among themselves in 5! ways
For circular table, the first position can be fixed, so effective arrangement is (5-1)! × 5! = 4! × 5! = 24 × 120 = 2,880
Short Trick: For alternating men/women in circular arrangement: (n/2-1)! × (n/2)! = 4! × 5! = 2,880