HEC USAT-E (Pre-Engineering) Mathematics: Permutation & Combination MCQs

Practice Permutation & Combination MCQs for HEC USAT-E (Pre-Engineering) Mathematics — topic-wise sets with solved answers.

HEC USAT-E (Pre-Engineering) Mathematics: Permutation & Combination MCQs — sample questions

  1. Question 1

    Q1. In how many ways can 5 boys and 3 girls be seated in a row such that no two girls are together?

    • A) 6! * 7C3
    • B) 5! * 6C3 * 3!
    • C) 6! * 7P3
    • D) 5! * 6P3

    Answer: 5! * 6C3 * 3!

    Explanation: First arrange 5 boys in 5! ways, then select 3 places out of 6 for girls in 6C3 ways, and arrange them in 3! ways.

  2. Question 2

    Q2. A bag contains 4 red and 6 black balls. One ball is drawn at random. What is the probability that it is red?

    • A) 4 / 10
    • B) 6 / 10
    • C) 2 / 5
    • D) Both A and C

    Answer: Both A and C

    Explanation: Total balls = 10, red balls = 4, probability = 4/10 = 2/5.

  3. Question 3

    Q3. The number of ways to choose 3 or more books from a shelf of 7 books is

    • A) 7C3 + 7C4 + 7C5 + 7C6 + 7C7
    • B) 2^7 - (7C0 + 7C1 + 7C2)
    • C) Both A and B
    • D) 7P3 + 7P4 + 7P5 + 7P6 + 7P7

    Answer: Both A and B

    Explanation: Both methods count the number of subsets with 3 or more elements: directly and by subtracting subsets with 0, 1, or 2 elements from total subsets.

  4. Question 4

    Q4. A committee of 5 is to be formed from 6 men and 4 women. In how many ways can this be done if at least 2 women are included?

    • A) 6C3 * 4C2 + 6C2 * 4C3 + 6C1 * 4C4
    • B) 4C2 * 6C3 + 4C3 * 6C2 + 4C4 * 6C1
    • C) Both A and B
    • D) 10C5 - (4C0 * 6C5 + 4C1 * 6C4)

    Answer: Both A and B

    Explanation: Count committees with 2, 3, or 4 women and corresponding men, or use the complementary counting principle.

  5. Question 5

    Q5. The probability that a leap year has 53 Sundays is

    • A) 1 / 7
    • B) 2 / 7
    • C) 3 / 7
    • D) 4 / 7

    Answer: 2 / 7

    Explanation: A leap year has 52 weeks and 2 extra days. For 53 Sundays, one of the extra days must be Sunday, probability = 2/7.

  6. Question 6

    Q6. In a class of 30 students, 15 are studying French, 12 are studying Spanish, and 5 are studying both. What is the probability that a randomly chosen student is studying exactly one of French or Spanish?

    • A) 3 / 5
    • B) 2 / 5
    • C) 7 / 15
    • D) 17/30

    Answer: 17/30

    Explanation: French only = 10 and Spanish only = 7, so exactly one language = 17 students out of 30.

  7. Question 7

    Q7. In how many ways can the letters of the word 'PERMUTATION' be arranged?

    • A) 11! / (2! * 2!)
    • B) 11! / 2!
    • C) 11!
    • D) 11! / (2! + 2!)

    Answer: 11! / 2!

    Explanation: PERMUTATION has 11 letters with T repeated twice, so arrangements = 11!/2!.

  8. Question 8

    Q8. A box contains 12 red, 8 blue, and 6 green balls. One ball is drawn at random. What is the probability that it is not blue?

    • A) 18 / 26
    • B) 8 / 26
    • C) 9 / 13
    • D) Both A and C

    Answer: Both A and C

    Explanation: Total balls = 26, not blue balls = 18, probability = 18/26 = 9/13.

  9. Question 9

    Q9. The number of ways to distribute 5 distinct balls into 3 distinct boxes is

    • A) 3^5
    • B) 5³
    • C) 3P5
    • D) 5C3 * 3!

    Answer: 3^5

    Explanation: Each ball has 3 choices of boxes, so total ways = 3 * 3 * 3 * 3 * 3 = 3^5.

  10. Question 10

    Q10. The probability of getting a sum of 7 when two dice are thrown is

    • A) 1 / 6
    • B) 1 / 3
    • C) 1 / 12
    • D) 1 / 9

    Answer: 1 / 6

    Explanation: Total outcomes = 6 * 6 = 36, favorable outcomes = 6 (1,6; 2,5; 3,4; 4,3; 5,2; 6,1), probability = 6/36 = 1/6.

  11. Question 11

    Q11. How many 3-digit numbers can be formed using the digits 1, 2, 3, 4, 5 without repetition?

    • A) 5P3
    • B) 5C3 * 3!
    • C) Both A and B
    • D) 5³

    Answer: Both A and B

    Explanation: Selecting 3 out of 5 digits and arranging them, so 5P3 = 5C3 * 3!.

  12. Question 12

    Q12. A coin is tossed 3 times. What is the probability of getting at least 2 heads?

    • A) 1 / 2
    • B) 3 / 8
    • C) 1 / 2 + 1 / 8
    • D) 4 / 8

    Answer: 1 / 2 + 1 / 8

    Explanation: Probability of exactly 2 heads + probability of 3 heads = 3C2 * (1/2)³ + (1/2)³ = 3/8 + 1/8 = 1/2.

  13. Question 13

    Q13. The number of diagonals in a polygon with 8 sides is

    • A) 8C2 - 8
    • B) 8 * (8 - 3) / 2
    • C) Both A and B
    • D) 8P2 - 8

    Answer: Both A and B

    Explanation: Total line segments between 8 vertices = 8C2, subtract the 8 sides to get diagonals, or use the formula n(n-3)/2.

  14. Question 14

    Q14. In how many ways can 4 people be seated around a circular table?

    • A) 4!
    • B) 3!
    • C) (4-1)!
    • D) Both B and C

    Answer: Both B and C

    Explanation: For circular permutations, fix one person and arrange the rest, so (n-1)! = 3!.

  15. Question 15

    Q15. A bag contains 5 white and 3 black balls. Two balls are drawn at random. What is the probability that they are of different colors?

    • A) 15 / 28
    • B) 15 / 56
    • C) 1 / 2
    • D) 13 / 28

    Answer: 15 / 28

    Explanation: Total ways to draw 2 balls = 8C2, ways to draw 1 white and 1 black = 5C1 * 3C1, probability = (5*3) / (8C2) = 15/28.

  16. Question 16

    Q16. The number of ways to choose a chairman and a vice-chairman from a committee of 8 people is

    • A) 8P2
    • B) 8C2 * 2!
    • C) Both A and B
    • D) 8²

    Answer: Both A and B

    Explanation: Selecting 2 people out of 8 and arranging them in 2 positions, so 8P2 = 8C2 * 2!.

  17. Question 17

    Q17. A die is rolled. What is the probability of getting a number greater than 4?

    • A) 1 / 3
    • B) 1 / 2
    • C) 2 / 3
    • D) 1 / 6

    Answer: 1 / 3

    Explanation: Favorable outcomes = 2 (5, 6), total outcomes = 6, probability = 2/6 = 1/3.

  18. Question 18

    Q18. In a group of 10 people, 5 are to be selected. What is the probability that a particular person is included?

    • A) 1 / 2
    • B) 5 / 10
    • C) Both A and B
    • D) 9C4 / 10C5

    Answer: Both A and B

    Explanation: If a particular person is to be included, select 4 more out of the remaining 9, probability = 9C4 / 10C5 = 1/2.

  19. Question 19

    Q19. The number of ways to distribute 4 identical items into 3 distinct boxes is

    • A) 3⁴
    • B) 4+3-1C3-1
    • C) 6C2
    • D) Both B and C

    Answer: Both B and C

    Explanation: Using stars and bars method, the number of ways = (n+k-1)C(k-1) = 6C2.

  20. Question 20

    Q20. A and B are two events such that P(A) = 1/3, P(B) = 1/2, and P(A ∩ B) = 1/6. What is P(A ∪ B)?

    • A) 2 / 3
    • B) 1 / 2
    • C) 3 / 4
    • D) 1 / 6

    Answer: 2 / 3

    Explanation: Use the formula P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 1/3 + 1/2 - 1/6 = 2/3.

Loading...