Air University Entry Test Mathematics Permutation & Combination — Set 3

Permutation & Combination MCQs set 3 for Air University Entry Test Mathematics — 20 solved questions.

Air University Entry Test Mathematics Permutation & Combination — Set 3

  1. Question 1

    Q1. A committee of 4 is to be formed from 5 men and 4 women. How many ways can this be done if at least one woman is included?

    • A) 120
    • B) 125
    • C) 115
    • D) 121

    Answer: 120

    Explanation: Total committees = 9C4, committees with no women = 5C4. Committees with at least one woman = 9C4 - 5C4 = 120

  2. Question 2

    Q2. In how many ways can 4 boys and 4 girls be seated in a row so that boys and girls are alternate?

    • A) 1152
    • B) 576
    • C) 2304
    • D) 1000

    Answer: 1152

    Explanation: First, arrange 4 boys in 4! ways, then 4 girls in 4! ways. Total = 4! * 4! * 2 = 1152

  3. Question 3

    Q3. Two dice are thrown. What is the probability of getting a sum of 7?

    • A) 1 / 6
    • B) 1 / 12
    • C) 1 / 3
    • D) 1 / 4

    Answer: 1 / 6

    Explanation: Favorable outcomes = 6. Total outcomes = 36. Probability = 6 / 36 = 1 / 6

  4. Question 4

    Q4. A bag contains 3 red, 4 blue, and 5 green balls. What is the probability of drawing a blue ball?

    • A) 1 / 3
    • B) 4 / 12
    • C) 1 / 4
    • D) 1 / 2

    Answer: 4 / 12

    Explanation: Total balls = 12, blue balls = 4. Probability = 4 / 12 = 1 / 3

  5. Question 5

    Q5. In how many ways can 6 people be divided into 3 groups of 2?

    • A) 15
    • B) 90
    • C) 45
    • D) 10

    Answer: 15

    Explanation: First, choose 2 out of 6, then 2 out of 4, and the last 2 are fixed. Total = 6C2 * 4C2 / 3! = 15

  6. Question 6

    Q6. A die is rolled twice. What is the probability of getting a sum of 9?

    • A) 1 / 9
    • B) 1 / 6
    • C) 1 / 12
    • D) 1 / 18

    Answer: 1 / 9

    Explanation: Favorable outcomes = 4. Total outcomes = 36. Probability = 4 / 36 = 1 / 9

  7. Question 7

    Q7. How many ways can the letters of the word 'MATHS' be arranged?

    • A) 120
    • B) 60
    • C) 24
    • D) 12

    Answer: 120

    Explanation: Total letters = 5, all distinct. Arrangements = 5! = 120

  8. Question 8

    Q8. A committee of 3 is to be formed from 4 men and 3 women. How many ways can this be done if at least one man is included?

    • A) 30
    • B) 35
    • C) 34
    • D) 31

    Answer: 34

    Explanation: Total committees = 7C3, committees with no men = 3C3. Committees with at least one man = 7C3 - 3C3 = 34

  9. Question 9

    Q9. In how many ways 5 boys and 3 girls can be seated around a round table if no two girls are together?

    • A) 4! * 5P3
    • B) 4! * 5C3
    • C) 5! * 4P3
    • D) 5! * 4C3

    Answer: 4! * 5P3

    Explanation: First, arrange 5 boys in (5-1)! = 4! ways. Then, place 3 girls in 5 gaps between boys in 5P3 ways.

  10. Question 10

    Q10. A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random, find the probability that 2 are red and 1 is blue.

    • A) (5C2 * 3C1) / 8C3
    • B) (5C2 + 3C1) / 8C3
    • C) (5P2 * 3P1) / 8P3
    • D) (5C2 * 3C1) / 8P3

    Answer: (5C2 * 3C1) / 8C3

    Explanation: Use combination formula to find the total number of ways to draw 3 balls and the number of ways to draw 2 red and 1 blue.

  11. Question 11

    Q11. The number of ways to distribute 5 distinct objects into 3 distinct boxes is

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

    Answer: 3^5

    Explanation: Each object has 3 choices of boxes, so 5 objects have 3 * 3 * 3 * 3 * 3 = 3^5 ways to be distributed.

  12. Question 12

    Q12. If the probability of event A is 1 / 3 and the probability of event B is 1 / 4, then P(A ∪ B) = ? (A and B are independent)

    • A) 1 / 3 + 1 / 4
    • B) 1 / 3 + 1 / 4 - 1 / 12
    • C) 1 / 3 + 1 / 4 + 1 / 12
    • D) 1 / 3 * 1 / 4

    Answer: 1 / 3 + 1 / 4 - 1 / 12

    Explanation: For independent events, P(A ∪ B) = P(A) + P(B) - P(A) * P(B) = 1 / 3 + 1 / 4 - (1 / 3) * (1 / 4).

  13. Question 13

    Q13. The number of 3-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 without repetition is

    • A) 5P3
    • B) 5C3 * 3!
    • C) 5 * 4 * 3
    • D) All of the above

    Answer: All of the above

    Explanation: All options represent the same value: 5P3 = 5C3 * 3! = 5 * 4 * 3 = 60.

  14. Question 14

    Q14. A committee of 3 is to be formed from 5 men and 4 women. Find the probability that it has at least 1 woman.

    • A) 1 - (5C3 / 9C3)
    • B) (4C1 * 5C2) / 9C3
    • C) (4C1 * 5C2 + 4C2 * 5C1 + 4C3) / 9C3
    • D) Both A and C

    Answer: Both A and C

    Explanation: P(at least 1 woman) = 1 - P(no women) or sum of probabilities of 1, 2, or 3 women.

  15. Question 15

    Q15. If P(A) = 1 / 2 and P(B) = 1 / 3, and P(A ∩ B) = 1 / 6, then P(A / B) = ?

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

    Answer: 1 / 2

    Explanation: P(A / B) = P(A ∩ B) / P(B) = (1 / 6) / (1 / 3) = 1 / 2.

  16. Question 16

    Q16. The number of diagonals in a hexagon is

    • A) 6C2 - 6
    • B) 6C2
    • C) 6P2 - 6
    • D) 6

    Answer: 6C2 - 6

    Explanation: Total line segments between 6 vertices = 6C2. Subtract 6 sides to get the number of diagonals.

  17. Question 17

    Q17. A coin is tossed 3 times. Find the probability of getting at least 2 heads.

    • A) (3C2 + 3C3) / 2³
    • B) 3C2 / 2³
    • C) (3C2 + 3C3) / 3!
    • D) 1 / 2

    Answer: (3C2 + 3C3) / 2³

    Explanation: P(at least 2 heads) = P(2 heads) + P(3 heads) = (3C2 + 3C3) / total outcomes.

  18. Question 18

    Q18. The number of ways to arrange the letters of the word 'LEDGER' is

    • A) 6! / 2!
    • B) 6!
    • C) 6! / 3!
    • D) 5!

    Answer: 6! / 2!

    Explanation: There are 6 letters with 'E' repeated twice, so the total arrangements = 6! / 2!.

  19. Question 19

    Q19. If the letters of the word 'ARTICLE' are arranged randomly, find the probability that vowels occupy even places.

    • A) (4C3 * 3! * 4!) / 7!
    • B) (4P3 * 4!) / 7!
    • C) (3! * 4!) / 7!
    • D) 4! / 7!

    Answer: (3! * 4!) / 7!

    Explanation: 3 vowels can be arranged in 4 even places in 4P3 = 4C3 * 3! ways. Remaining 4 consonants can be arranged in 4! ways.

  20. Question 20

    Q20. A box contains 4 red, 5 white, and 6 blue balls. If 3 balls are drawn at random, find the probability that they are of different colors.

    • A) (4C1 * 5C1 * 6C1) / 15C3
    • B) (4 + 5 + 6)C3 / 15C3
    • C) (4C1 + 5C1 + 6C1) / 15C3
    • D) 4! * 5! * 6! / 15!

    Answer: (4C1 * 5C1 * 6C1) / 15C3

    Explanation: Choose 1 ball from each color in (4C1 * 5C1 * 6C1) ways. Total ways to draw 3 balls = 15C3.

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Level 1

A committee of 4 is to be formed from 5 men and 4 women. How many ways can this be done if at least one woman is included?